Similar Triangles

Two triangles are similar if they have the same shape: the angles of one triangle are equal to the angles of the other, and the sides of one triangle are proportional to the corresponding sides of the other.

One of two similar triangles is an enlarged or reduced copy of the other, which may also be moved, turned or flipped over. A photograph and its enlargement, a map and the real terrain, two objects and their shadows at the same moment: all of them give similar triangles.

Definition and Notation

Triangles $ABC$ and $A_1B_1C_1$ are similar if

We write ${\triangle ABC\sim\triangle A_1B_1C_1}$ and read it "triangle $ABC$ is similar to triangle $A_1B_1C_1$". The number $k$ is the scale factor (the ratio of similarity). With the usual notation for the sides, ${a_1=ka},$ ${b_1=kb}$ and ${c_1=kc}.$

• The order of the letters matters. It shows which vertices correspond: $A$ to $A_1,$ $B$ to $B_1$ and $C$ to $C_1.$ Corresponding sides join corresponding vertices, and they lie opposite equal angles.
• If ${k\gt 1},$ the second triangle is an enlargement of the first; if ${k\lt 1},$ it is a reduction. In the figure above ${k=1.55}.$
• If ${k=1},$ the triangles are congruent. So congruent triangles are similar, but similar triangles are usually not congruent.
• If ${\triangle ABC\sim\triangle A_1B_1C_1}$ with the scale factor $k,$ then ${\triangle A_1B_1C_1\sim\triangle ABC}$ with the scale factor $\frac1k.$ Two triangles that are similar to a third triangle are similar to each other.

When only the side lengths are known, the corresponding sides are found by their size: the shortest side corresponds to the shortest one, the middle one to the middle one and the longest one to the longest one.

On this page: Criteria · Right and isosceles triangles · A line parallel to a side · Perimeters and areas · In a right triangle · Menelaus · Ceva · Solved problems · Real-life problems · Common mistakes · Practice

Similarity Criteria

To prove that two triangles are similar, we do not need all three angles and all three ratios of sides. Each of the following criteria is enough. In the criteria ${k\gt 0}$ is any number.

Angle-Angle (AA)

If two angles of one triangle are equal to two angles of another triangle, the triangles are similar.

The third angles are then equal too, because each of them is $180^\circ$ minus the other two. This is the criterion used most often: parallel lines, vertical angles, a common angle and right angles all give equal angles.

Side-Angle-Side (SAS)

If two sides of one triangle are proportional to two sides of another triangle and the angles between these sides are equal, the triangles are similar.

If ${A_1B_1=k\cdot AB},$ ${A_1C_1=k\cdot AC}$ and ${\angle A=\angle A_1},$ then ${\triangle ABC\sim\triangle A_1B_1C_1}.$ The equal angle must be the one between the two sides.

Side-Side-Side (SSS)

If the three sides of one triangle are proportional to the three sides of another triangle, the triangles are similar.

To check it, write both sets of sides in increasing order and divide: the three quotients must be equal.

Two sides and the angle opposite the longer one

If two sides of one triangle are proportional to two sides of another triangle and the angles opposite the longer of these sides are equal, the triangles are similar.

If the equal angles lie opposite the shorter sides, the triangles may not be similar. For example, the triangles with sides 6, 8, 10 and 6, 8, 2.8 both have the sides 6 and 8 and the same angle opposite the side 6, but the first one is a right triangle and the second one is obtuse.

Why the criteria are true

We use the theorem from the section A line parallel to a side: such a line cuts off a triangle similar to the given one. Let ${A_1B_1=k\cdot AB}$ with ${k\gt 1}$ (if ${k\lt 1},$ swap the triangles; if ${k=1},$ the criteria are the congruence criteria). On the side $A_1B_1$ mark the point $D$ with ${A_1D=AB}$ and draw ${DE\parallel B_1C_1}$ with $E$ on $A_1C_1.$ Then ${\triangle A_1DE\sim\triangle A_1B_1C_1}$ with the scale factor $\frac1k,$ so ${A_1E=\frac{A_1C_1}{k}},$ ${DE=\frac{B_1C_1}{k}}$ and ${\angle A_1DE=\angle B_1}.$

• AA. ${A_1D=AB},$ ${\angle A_1=\angle A}$ and ${\angle A_1DE=\angle B_1=\angle B},$ so ${\triangle A_1DE\cong\triangle ABC}$ (ASA).
• SAS. ${A_1E=\frac{A_1C_1}{k}=AC},$ so ${\triangle A_1DE\cong\triangle ABC}$ (SAS).
• SSS. ${A_1E=\frac{A_1C_1}{k}=AC}$ and ${DE=\frac{B_1C_1}{k}=BC},$ so ${\triangle A_1DE\cong\triangle ABC}$ (SSS).
• The fourth criterion works in the same way with the congruence criterion "two sides and the angle opposite the longer one": label the triangles so that ${AB\ge AC}$ and ${\angle C=\angle C_1};$ then ${A_1E=AC}$ and ${\angle A_1ED=\angle C_1=\angle C}.$

In each case $ABC$ is congruent to $A_1DE,$ which is similar to $A_1B_1C_1,$ so ${\triangle ABC\sim\triangle A_1B_1C_1}.$

Right, isosceles and equilateral triangles

For these triangles the criteria become shorter.

• Two right triangles are similar if an acute angle of one equals an acute angle of the other (AA), if their legs are proportional (SAS with the right angle), or if a leg and the hypotenuse of one are proportional to a leg and the hypotenuse of the other (the hypotenuse is the longer side and lies opposite the right angle).
• Two isosceles triangles are similar if their vertex angles are equal, if their base angles are equal, or if the base and the leg of one are proportional to the base and the leg of the other.
• Any two equilateral triangles are similar: all their angles are $60^\circ.$

A Line Parallel to a Side

Why? The angle $A$ is common, and ${\angle ADE=\angle ABC}$ as corresponding angles at the parallel lines $DE$ and $BC$ (and so are ${\angle AED=\angle ACB}$). So the two triangles have equal angles. The ratios ${\frac{AD}{AB}=\frac{AE}{AC}}$ follow from the intercept theorem. For the third ratio draw through $E$ the line parallel to $AB;$ it meets $BC$ at a point $F.$ Then $DBFE$ is a parallelogram, ${DE=BF},$ and by the intercept theorem for the angle $C$ we get ${\frac{BF}{BC}=\frac{AE}{AC}}.$

The ratio of the parallel sides is ${\frac{DE}{BC}=\frac{AD}{AB}},$ with the whole side $AB$ in the denominator. The ratio ${\frac{AD}{DB}}$ equals ${\frac{AE}{EC}},$ but it is not equal to ${\frac{DE}{BC}}.$

The same holds when the line crosses the extensions of the sides beyond the vertex $A\text{:}$ the triangles $ADE$ and $ABC$ form an "hourglass" and are similar by AA (vertical angles at $A$ and alternate angles at the parallel lines).

The converse is also true: if $D$ lies on $AB,$ $E$ lies on $AC$ and ${\frac{AD}{AB}=\frac{AE}{AC}},$ then ${DE\parallel BC}.$ Indeed, the triangles $ADE$ and $ABC$ are similar by SAS, so ${\angle ADE=\angle ABC},$ and these are corresponding angles.

Altitudes, Perimeters and Areas of Similar Triangles

Proof

Altitudes. The triangles $ACH$ and $A_1C_1H_1$ have right angles at $H$ and $H_1$ and equal angles at $A$ and $A_1$ (we may take $\angle A$ acute, because at most one angle of a triangle is not acute; then ${\angle CAH=\angle A}$ even when $H$ lies beyond $B$). By AA they are similar, so ${\frac{C_1H_1}{CH}=\frac{A_1C_1}{AC}=k}.$

Medians. If $M$ and $M_1$ are the midpoints of $AB$ and $A_1B_1,$ then ${A_1M_1=k\cdot AM},$ ${A_1C_1=k\cdot AC}$ and ${\angle A_1=\angle A},$ so ${\triangle AMC\sim\triangle A_1M_1C_1}$ by SAS and ${C_1M_1=k\cdot CM}.$ For the angle bisectors $CL$ and $C_1L_1$ use the triangles $ACL$ and $A_1C_1L_1:$ they are similar by AA, because ${\angle A=\angle A_1}$ and ${\angle ACL}={\frac12\angle C}={\frac12\angle C_1}={\angle A_1C_1L_1},$ so ${C_1L_1=k\cdot CL}.$

Perimeters. ${P_1}={a_1+b_1+c_1}={ka+kb+kc}={kP}.$

Areas. ${S_1}={\frac12c_1h_{c_1}}={\frac12\cdot kc\cdot kh_c}={k^2\cdot\frac12ch_c}={k^2S}.$

Radii. With the semiperimeter $p,$ ${r=\frac{S}{p}}$ and ${R=\frac{abc}{4S}}$ (see Problem 10). The area is multiplied by $k^2$ and $p$ by $k,$ so $r$ is multiplied by $k;$ $abc$ is multiplied by $k^3,$ so $R$ is multiplied by $k.$

For example, if every side is doubled, the perimeter is doubled, but the area becomes four times larger. On a map with the scale ${1:5000}$ the lengths are reduced 5000 times, and the areas ${5000^2=25\,000\,000}$ times.

Similar Triangles in a Right Triangle

Let ${\angle C=90^\circ}$ and let $CH$ be the altitude to the hypotenuse. It divides the hypotenuse $c$ into the segments ${AH=m}$ and ${HB=n},$ the projections of the legs $b$ and $a.$

The altitude to the hypotenuse divides a right triangle into two triangles that are similar to it and to each other: ${\triangle ACH\sim\triangle ABC\sim\triangle CBH}.$

Each small triangle has a right angle and shares an acute angle with the big one ($\angle A$ and $\angle B$), so the criterion AA applies. Writing the proportions of the corresponding sides, we get

For example, from ${\triangle ACH\sim\triangle CBH}$ we get ${\frac{AH}{CH}=\frac{CH}{HB}},$ that is ${h^2=mn}.$ Adding the last two equalities gives ${a^2+b^2=c(m+n)=c^2}:$ this is a proof of the Pythagorean theorem. More about these formulas is on the page Altitude of a triangle.

Famous Theorems: Menelaus, Ceva and Morley

Menelaus's theorem

The product goes around the triangle: from $A$ to the point on $AC,$ then to $C,$ to the point on $CB,$ to $B,$ to the point on the line $AB$ and back to $A.$ The theorem also holds when the line crosses the extensions of all three sides.

Proof

Draw through $C$ the line parallel to $AB;$ it meets the line $MN$ at $Q.$ The triangles $MCQ$ and $MAP$ have equal vertical angles at $M$ and equal alternate angles at $C$ and $A,$ so they are similar and ${\frac{AM}{MC}=\frac{AP}{CQ}}.$ In the same way ${\triangle NCQ\sim\triangle NBP},$ so ${\frac{CN}{NB}=\frac{CQ}{BP}}.$ Then

$$\frac{AM}{MC}\cdot\frac{CN}{NB}\cdot\frac{BP}{PA}=\frac{AP}{CQ}\cdot\frac{CQ}{BP}\cdot\frac{BP}{PA}=1.$$

The converse is also true: if $M$ and $N$ lie on the sides $AC$ and $BC,$ $P$ lies on the extension of $AB$ and the product equals 1, then $M,$ $N$ and $P$ lie on one line. So Menelaus's theorem is used to prove that three points lie on one line.

Ceva's theorem

Proof

Draw through $A$ the line parallel to $BC.$ The extensions of $BB_1$ and $CC_1$ meet it at $X$ and $Y.$ Vertical angles and alternate angles at parallel lines give four pairs of similar triangles:

• ${\triangle AC_1Y\sim\triangle BC_1C},$ so ${\frac{AC_1}{C_1B}=\frac{AY}{BC}};$
• ${\triangle CB_1B\sim\triangle AB_1X},$ so ${\frac{CB_1}{B_1A}=\frac{BC}{AX}};$
• ${\triangle OA_1B\sim\triangle OAX}$ and ${\triangle OA_1C\sim\triangle OAY},$ so ${\frac{BA_1}{AX}=\frac{OA_1}{OA}=\frac{A_1C}{AY}},$ that is ${\frac{BA_1}{A_1C}=\frac{AX}{AY}}.$

Multiplying, we get ${\frac{AY}{BC}\cdot\frac{AX}{AY}\cdot\frac{BC}{AX}=1}.$

The converse is also true. For the medians each ratio equals 1, so the product is 1, and the converse shows that the three medians meet at one point.

Morley's theorem

Trisect each angle of a triangle, that is, divide it into three equal angles. The trisectors next to each side meet at a point, and the three points obtained are always the vertices of an equilateral triangle. This surprising theorem was discovered by Frank Morley in 1899; its proof is harder and is not based only on similar triangles.

Solved Problems

Problem 1. Are the triangles $ABC$ and $PQR$ in the figure similar?

Solution. We know all six sides, so we use SSS. In increasing order the sides are 2, 4, 5 and 6, 12, 15:

$$\frac{PQ}{AB}=\frac{6}{2}=3,\qquad \frac{QR}{BC}=\frac{12}{4}=3,\qquad \frac{PR}{AC}=\frac{15}{5}=3.$$

The three ratios are equal, so ${\triangle ABC\sim\triangle PQR}$ with the scale factor 3.

Problem 2. In the triangles $ABC$ and $PQR$ we have ${\angle A=\angle P},$ ${\angle B=\angle Q},$ ${AB=4},$ ${BC=6},$ ${AC=7}$ and ${QR=12}.$ Find $PQ$ and $PR.$

Solution. By AA ${\triangle ABC\sim\triangle PQR}.$ The triangle $PQR$ is drawn flipped over, but the equal angles give the correspondence: $A$ to $P,$ $B$ to $Q$ and $C$ to $R.$ So $BC$ corresponds to $QR$ and ${k=\frac{QR}{BC}=\frac{12}{6}=2}.$ Then ${PQ=2\cdot AB=8}$ and ${PR=2\cdot AC=14}.$

Problem 3. The points $B$ and $C$ lie on the sides $AD$ and $AE$ of triangle $ADE,$ ${BC\parallel DE},$ ${BC=3},$ ${DE=6}$ and ${BD=4}.$ Find $AB.$

Solution. The line $BC$ is parallel to the side $DE,$ so ${\triangle ABC\sim\triangle ADE}$ and ${\frac{AB}{AD}=\frac{BC}{DE}=\frac36=\frac12}.$ With ${AD=AB+4}$ this gives ${2AB=AB+4},$ so ${AB=4}.$

Problem 4. The diagonals of the trapezoid $ABCD$ ${(AB\parallel CD)}$ meet at $O,$ and ${AO:OC=3:2}.$ The midline of the trapezoid is 5 cm. Find the bases.

Solution. In the triangles $AOB$ and $COD$ the angles at $O$ are vertical, and ${\angle OAB=\angle OCD}$ as alternate angles at the parallel lines $AB$ and $CD.$ So ${\triangle AOB\sim\triangle COD}$ and ${\frac{AB}{CD}=\frac{AO}{OC}=\frac32}.$ The midline is half the sum of the bases, so ${AB+CD=10}.$ Then ${AB=6}$ cm and ${CD=4}$ cm.

Problem 5. The sides of a triangle are in the ratio ${2:3:4}.$ The perimeter of a similar triangle is 84.6 cm. Find its sides.

Solution. The sides of the similar triangle are in the same ratio, so they are $2t,$ $3t$ and $4t.$ Then ${9t=84.6},$ ${t=9.4},$ and the sides are 18.8 cm, 28.2 cm and 37.6 cm.

Problem 6. A line parallel to the side $BC$ of triangle $ABC$ divides it into two parts of equal area. In what ratio does it divide the side $AB?$

Solution. Let the line meet $AB$ at $D$ and $AC$ at $E.$ Then ${\triangle ADE\sim\triangle ABC}$ with ${k=\frac{AD}{AB}},$ and the area of $ADE$ is half the area of $ABC.$ So ${k^2=\frac12},$ ${k=\frac{1}{\sqrt2}}$ and

$$\frac{AD}{DB}=\frac{AD}{AB-AD}=\frac{1}{\sqrt2-1}=\sqrt2+1\approx 2.41.$$

The line is not through the midpoint of $AB\text{:}$ the midline cuts off a triangle with only a quarter of the area.

Problem 7. In triangle $ABC$ the bisector $AD$ is drawn, and through $D$ the line parallel to $AB$ meets $AC$ at $E.$ Find $AB$ if ${AE=6}$ cm and ${EC=4}$ cm.

Solution. ${\angle ADE=\angle DAB}$ as alternate angles at the parallel lines $DE$ and $AB,$ and ${\angle DAB=\angle DAE}$ because $AD$ is the bisector. So the triangle $ADE$ is isosceles and ${DE=AE=6}.$ The line $DE$ is parallel to the side $AB,$ so ${\triangle EDC\sim\triangle ABC}$ and ${\frac{DE}{AB}=\frac{EC}{AC}=\frac{4}{10}}.$ Then ${AB=\frac{6\cdot 10}{4}=15}$ cm.

Problem 8. A square stands on the side ${AB=a}$ of triangle $ABC,$ and its other two vertices lie on $AC$ and $BC.$ Find the side $x$ of the square if ${a=15}$ cm and the altitude from $C$ is ${h=10}$ cm.

Solution. The upper side of the square is parallel to $AB,$ so it cuts off a triangle similar to $ABC.$ Its base is $x$ and its altitude is ${h-x}.$ The altitudes are in the same ratio as the bases:

$$\frac{x}{a}=\frac{h-x}{h}\quad\Rightarrow\quad x=\frac{ah}{a+h}=\frac{15\cdot 10}{25}=6\ \text{cm}.$$

Problem 9. The chords $AB$ and $CD$ of a circle meet at $P.$ Prove that ${PA\cdot PB=PC\cdot PD}$ and find $PD$ if ${PA=4},$ ${PB=6}$ and ${PC=3}.$

Solution. The inscribed angles $CAB$ and $CDB$ stand on the same arc $CB,$ so they are equal. The angles $APC$ and $DPB$ are vertical. By AA ${\triangle APC\sim\triangle DPB},$ hence ${\frac{PA}{PD}=\frac{PC}{PB}}$ and ${PA\cdot PB=PC\cdot PD}.$ Then ${PD=\frac{4\cdot 6}{3}=8}.$

Harder Problems

Problem 10. Prove that the product of two sides of a triangle equals the altitude to the third side times the diameter of the circumscribed circle: ${ab=2R\cdot h_c}.$

Solution. At most one angle of a triangle is not acute, so we may assume that the angle $B$ is acute (otherwise swap $A$ and $B$). Let $CH$ be the altitude and $CD$ the diameter of the circumscribed circle; $H$ lies on the ray $BA.$ The angle $CAD$ is inscribed in a semicircle, so ${\angle CAD=90^\circ=\angle CHB}.$ The points $B$ and $D$ lie on the same arc $CA\text{:}$ otherwise ${\angle CDA=180^\circ-\angle B}$ would be obtuse, but it is an acute angle of the right triangle $CAD.$ So the inscribed angles $CBA$ and $CDA$ stand on the same arc and are equal. By AA ${\triangle CHB\sim\triangle CAD},$ hence ${\frac{CH}{CA}=\frac{CB}{CD}},$ that is ${CA\cdot CB=CD\cdot CH},$ or ${ab=2R\cdot h_c}.$

Since ${h_c=\frac{2S}{c}},$ this gives the well-known formula ${S=\frac{abc}{4R}}.$

Problem 11. In the acute triangle $ABC$ the altitudes $CD$ and $AE$ are drawn. Prove that the segment $DE$ cuts off a triangle similar to $ABC.$

Solution. The right triangles $ABE$ and $CBD$ have the common angle $B,$ so they are similar and ${\frac{BE}{BD}=\frac{AB}{CB}},$ or ${\frac{BD}{BC}=\frac{BE}{BA}}.$ The triangles $BDE$ and $BCA$ have the common angle $B$ and these proportional sides, so by SAS ${\triangle BDE\sim\triangle BCA}$ with the scale factor ${\frac{BD}{BC}=\cos B}.$ Note the correspondence: $D$ corresponds to $C$ and $E$ to $A,$ so in general $DE$ is not parallel to $AC$ (it is only when ${AB=BC}$).

Problem 12 (the Euler line). Let $O$ be the center of the circumscribed circle, $G$ the centroid and $H$ the orthocenter (the intersection of the altitudes) of the acute triangle $ABC.$ Prove that $O,$ $G$ and $H$ lie on one line and ${GH=2\,GO}.$

Solution. In an equilateral triangle the three points coincide, so let the triangle be not equilateral and name its vertices so that ${AB\ne AC};$ then neither $H$ nor $O$ lies on the median from $A.$ Let $M$ and $N$ be the midpoints of $BC$ and $CA.$ The lines $OM$ and $AH$ are both perpendicular to $BC,$ so they are parallel; in the same way ${ON\parallel BH},$ and ${MN\parallel AB}$ as a midline. The triangles $OMN$ and $HAB$ have parallel sides, so their angles are equal and they are similar with the scale factor ${\frac{MN}{AB}=\frac12}.$ Hence ${AH=2\,OM}.$

The centroid lies on the median $AM$ with ${AG=2\,GM}.$ The triangle is acute, so $H$ and $O$ are inside it: the ray $AH$ goes towards $BC,$ and the ray $MO$ goes away from $BC.$ Hence $H$ and $O$ lie on different sides of $AM,$ and in the triangles $AGH$ and $MGO$ the angles $HAG$ and $OMG$ are equal as alternate angles at the parallel lines $AH$ and $OM,$ and ${\frac{AH}{MO}=\frac{AG}{GM}=2}.$ By SAS ${\triangle AGH\sim\triangle MGO},$ so ${\angle AGH=\angle MGO}.$ These angles lie on different sides of the line $AM,$ so $H,$ $G$ and $O$ lie on one line, and ${GH=2\,GO}.$

The same is true in every triangle that is not equilateral; this line is called the Euler line.

Real-Life Problems

Problem 13 (height from a shadow). At the same moment a person 1.8 m tall casts a shadow 2.7 m long, and a tree casts a shadow 15 m long. How tall is the tree?

Solution. Let $AB$ be the tree and $BC$ its shadow, $A_1B_1$ the person and $B_1C_1$ the person's shadow. The sun rays are parallel, so they make equal angles with the ground: ${\angle C=\angle C_1}.$ The tree and the person are vertical, so ${\angle B=\angle B_1=90^\circ}.$ By AA ${\triangle ABC\sim\triangle A_1B_1C_1}$ and

$$\frac{AB}{A_1B_1}=\frac{BC}{B_1C_1}\quad\Rightarrow\quad AB=\frac{1.8\cdot 15}{2.7}=10\ \text{m}.$$

Problem 14 (sighting over a tree). Trisha wants to find the height of a building. A tree 2.8 m tall stands 30 m in front of the building. Trisha moves back from the tree until she sees the top of the tree and the top of the building in one line; then she is 5 m from the tree. Her eyes are 1.6 m above the ground. How tall is the building?

Solution. Let $T$ be the top of the tree and $D$ the top of the building. Draw the horizontal line through her eye $E;$ it meets the tree at $K$ and the building at $L.$ The right triangles $EKT$ and $ELD$ have the common angle at $E,$ so they are similar. Here ${EK=5},$ ${EL=5+30=35}$ and ${KT=2.8-1.6=1.2}$ m, so

$$\frac{LD}{KT}=\frac{EL}{EK}\quad\Rightarrow\quad LD=\frac{1.2\cdot 35}{5}=8.4\ \text{m}.$$

The building is ${8.4+1.6=10}$ m tall.

Problem 15 (the width of a river). To find the distance from the point $A$ to the point $X$ on the other bank, we walk from $A$ along the bank perpendicular to $AX$ and mark the point $B$ with ${AB=200}$ m. On $AB$ we mark $M$ with ${MB=30}$ m, walk from $M$ parallel to $AX$ and stop at the point $N$ that lies on the line $BX.$ We measure ${MN=15}$ m. Find $AX.$

Solution. ${MN\parallel AX},$ so ${\triangle BMN\sim\triangle BAX}$ and

$$\frac{AX}{MN}=\frac{AB}{MB}\quad\Rightarrow\quad AX=\frac{15\cdot 200}{30}=100\ \text{m}.$$

Problem 16 (a conveyor belt). An inclined conveyor belt starts at the point $A$ on the floor and rests on a post 3 m high that stands 8 m from $A.$ The belt must be extended at the same slope to a new post 9 m high. How far behind the first post does the new post stand, and how much longer does the belt become?

Solution. Let $D$ and $E$ be the foot and the top of the first post, $B$ and $C$ those of the new post, and ${x=DB}.$ The right triangles $ADE$ and $ABC$ have the common angle $A,$ so they are similar with ${k=\frac{BC}{DE}=\frac93=3}.$ Then ${AB=3\cdot 8=24}$ m and ${x=24-8=16}$ m. The old belt is ${AE}={\sqrt{8^2+3^2}}={\sqrt{73}}\approx{8.54}$ m. The new one is ${AC=3\cdot AE},$ so the belt becomes longer by ${EC=2\sqrt{73}\approx 17.09}$ m.

Problem 17 (area on a map). On a map with the scale ${1:5000}$ a triangular field has an area of 12 cm². What is its real area?

Solution. The field and its picture are similar with ${k=5000},$ so the real area is ${12\cdot 5000^2}={300\,000\,000}$ cm² ${=30\,000}$ m², that is 3 hectares.

Common Mistakes

Mistake	Wrong	Right
Sides that do not correspond	${\triangle ABC\sim\triangle PQR},$ so ${\frac{AB}{QR}=\frac{BC}{PQ}}$	Pair the sides by the order of the letters: ${\frac{AB}{PQ}=\frac{BC}{QR}=\frac{AC}{PR}}$
A part of the side instead of the whole side	${DE\parallel BC},$ so ${\frac{DE}{BC}=\frac{AD}{DB}}$	${\frac{DE}{BC}=\frac{AD}{AB}}$
Areas scaled like lengths	The sides are doubled, so the area is doubled	The area is multiplied by ${2^2=4}$
The angle is not between the sides	Two sides are proportional and some angle is equal, so the triangles are similar	The angle must be between the sides (SAS) or opposite the longer of them
Adding instead of multiplying	3, 4, 5 and 4, 5, 6 are similar: each side is 1 longer	Similarity needs equal ratios: ${\frac43\ne\frac54},$ so they are not similar
Polygons treated like triangles	The sides of a square and of a rhombus are proportional, so they are similar	Only for triangles do proportional sides give similarity; a rhombus with a ${60^\circ}$ angle is not similar to a square

Practice Problems

Try them yourself, then open the answer.

1. The sides of a triangle are 26 cm, 38 cm and 46 cm. The shortest side of a similar triangle is 13 cm. Find its other sides.

${k=\frac{13}{26}=\frac12},$ so the other sides are 19 cm and 23 cm.

2. Two corresponding sides of similar triangles are 15 cm and 6 cm, and the altitude drawn to the side of 15 cm is 8 cm. Find the corresponding altitude drawn to the side of 6 cm.

${8\cdot\frac{6}{15}=3.2}$ cm.

3. The scale factor of two similar triangles is 1.75, and the sum of two corresponding altitudes is 121 mm. Find these altitudes.

${h+1.75h=121},$ so ${h=44}$ mm and ${1.75h=77}$ mm.

4. The ratio of two corresponding altitudes of similar triangles is 0.6, and two corresponding angle bisectors differ by 1.4 cm. Find these bisectors.

The bisectors are in the same ratio: ${l-0.6l=1.4},$ so ${l=3.5}$ cm and ${0.6l=2.1}$ cm.

5. The areas of two similar triangles are 18 cm² and 50 cm², and a side of the smaller one is 6 cm. Find the corresponding side of the larger one.

${k^2=\frac{50}{18}=\frac{25}{9}},$ ${k=\frac53},$ so the side is 10 cm.

6. The vertex angle of one isosceles triangle is $\alpha,$ and a base angle of another isosceles triangle is ${90^\circ-\frac{\alpha}{2}}.$ The leg of the first triangle and the perimeter of the second one are 80 cm each, and the base of the second one is 20 cm. Find the base of the first triangle.

The vertex angle of the second triangle is ${180^\circ-2\left(90^\circ-\frac{\alpha}{2}\right)=\alpha},$ so the triangles are similar. The legs of the second one are 30 cm, so the base of the first is ${80\cdot\frac{20}{30}=53\frac13}$ cm.

7. The base of an isosceles triangle is 3 cm and its leg is 6 cm. A line parallel to the base cuts off a segment between the legs equal to the part of the leg next to the base. Find this segment.

If the segment is $x,$ the small triangle has the leg ${6-x}$ and the base $x,$ so ${\frac{x}{3}=\frac{6-x}{6}}$ and ${x=2}$ cm.

8. In triangle $ABC$ ${AB=9},$ ${BC=12}$ and ${AC=6}.$ The point $D$ lies on $AB$ with ${AD=4}.$ Find $CD.$

${\frac{AD}{AC}=\frac{AC}{AB}=\frac23}$ and the angle $A$ is common, so ${\triangle ADC\sim\triangle ACB}$ (SAS) and ${CD=\frac23\cdot BC=8}.$

9. The parallelogram $AKMP$ is inscribed in triangle $ABC$ with the common angle $A$ ($K$ on $AC,$ $M$ on $BC,$ $P$ on $AB$). Find its sides if ${AC=24},$ ${AB=36}$ and ${KM:MP=3:1}.$

Let ${MP=AK=t},$ ${KM=3t}.$ ${KM\parallel AB},$ so ${\frac{3t}{36}=\frac{24-t}{24}},$ which gives ${t=8}.$ The sides are 8 and 24.

10. The legs of a trapezoid are 12 cm and 8 cm and its longer base is 27 cm. A diagonal divides the trapezoid into two similar triangles. Find this diagonal and the shorter base.

Let ${AB=27}$ and let $AC$ be the diagonal. ${\angle DCA=\angle CAB}$ as alternate angles. If also ${\angle D=\angle B},$ then ${\angle DAC=\angle ACB},$ so ${AD\parallel BC}$ and $ABCD$ would be a parallelogram. Hence ${\angle D=\angle ACB},$ ${\triangle DCA\sim\triangle CAB}$ and ${\frac{DC}{CA}=\frac{CA}{AB}=\frac{AD}{BC}}.$ The square of this ratio is ${\frac{DC}{CA}\cdot\frac{CA}{AB}=\frac{DC}{AB}\lt 1},$ so ${AD=8},$ ${BC=12}$ and the ratio is $\frac23.$ Then ${CA=\frac23\cdot 27=18}$ cm and ${DC=\frac23\cdot 18=12}$ cm.

11. In the rectangle $ABCD$ ${AB=10}$ and ${AD=4}.$ From a point $M$ on $CD$ the side $AB$ is seen at a right angle. Find $DM$ and $MC.$

${\triangle ADM\sim\triangle MCB}$ (right angles at $D$ and $C,$ and ${\angle DAM}={\angle CMB}={90^\circ-\angle DMA}$), so ${DM\cdot MC=4\cdot 4=16}$ and ${DM+MC=10}.$ So ${DM=2},$ ${MC=8}$ or ${DM=8},$ ${MC=2}.$

12. A right triangle with a leg of 8 cm is inscribed in a circle of radius 5 cm. Find the projection of this leg on the hypotenuse.

The hypotenuse is a diameter, 10 cm. From ${b^2=c\cdot m}$ we get ${m=\frac{64}{10}=6.4}$ cm.

13. The base of an isosceles triangle is 6 cm and its leg is 9 cm. Find the distance between the points where the inscribed circle touches the legs.

The tangent segment from the vertex opposite the base to the circle is ${9-3=6}$ cm. The segment is parallel to the base, so it is ${6\cdot\frac69=4}$ cm.

14. The base of an isosceles triangle is 12 cm and its leg is 9 cm. The tangent to the inscribed circle parallel to the base is drawn. Find the part of this tangent between the legs.

${h=\sqrt{81-36}=3\sqrt5},$ ${r=\frac{S}{p}=\frac{18\sqrt5}{15}=1.2\sqrt5}.$ The small triangle has the altitude ${h-2r=0.6\sqrt5=\frac{h}{5}},$ so the segment is ${\frac{12}{5}=2.4}$ cm.

15. The circles with radii 36 cm and 24 cm lie outside each other. Their common internal tangent divides the segment between the centers into parts that differ by 16 cm. Find the distance between the centers.

The two right triangles formed by the radii to the points of tangency are similar, so the parts are in the ratio ${36:24=3:2}.$ Then the parts are 48 cm and 32 cm, and the distance is 80 cm.

16. Two circles with radii 3 cm and 2 cm touch each other from outside. Their common external tangents meet at $M.$ Find the distance from the center of the smaller circle to $M.$

If the distance is $d,$ then ${\frac{d}{d+5}=\frac23}$ (similar right triangles with the radii to a tangent), so ${d=10}$ cm.

17. The bases $AB$ and $CD$ of the trapezoid $ABCD$ are in the ratio ${5:1},$ and ${AC=22}$ cm. Into what parts does the line through $B$ and the midpoint of $AD$ divide $AC?$

The line meets $CD$ extended at $K$ with ${DK=AB}$ (the triangles with the vertex at the midpoint of $AD$ are congruent). If the line meets $AC$ at $O,$ then ${\triangle AOB\sim\triangle COK}$ with ${\frac{AB}{CK}=\frac{5}{6}}.$ So ${AO=10}$ cm and ${OC=12}$ cm.

18. The median $BM$ of triangle $ABC$ makes with $AB$ an angle equal to the angle $C.$ Prove that ${AC^2=2AB^2}.$

${\triangle ABM\sim\triangle ACB}$ (AA: the angle $A$ is common and ${\angle ABM=\angle C}$). So ${\frac{AB}{AC}=\frac{AM}{AB}},$ ${AB^2=AC\cdot\frac{AC}{2}},$ and ${AC^2=2AB^2}.$

19. From a point $P$ outside a circle the tangent $PB$ and a line meeting the circle at $C$ and $A$ (in this order) are drawn. Prove that ${PB^2=PC\cdot PA}.$

The angle between the tangent and the chord $BC$ equals the inscribed angle $BAC,$ so ${\angle PBC=\angle PAB}.$ The angle $P$ is common, so ${\triangle PBC\sim\triangle PAB}$ and ${\frac{PB}{PA}=\frac{PC}{PB}},$ that is ${PB^2=PC\cdot PA}.$

20. The altitudes $BD$ and $CE$ of the acute triangle $ABC$ meet at $H.$ Prove that ${EH\cdot HC=DH\cdot HB}.$

The right triangles $EHB$ and $DHC$ have equal vertical angles at $H,$ so they are similar and ${\frac{EH}{DH}=\frac{HB}{HC}}.$

More about triangles: Triangles · Congruence of triangles · Altitude · Median · Intercept theorem problems