Similar Triangles: Problems with Solutions

Solution:
By the angle bisector theorem, the bisector divides the opposite side in the ratio of the adjacent sides: [tex]\frac{CL}{BL}=\frac{AC}{AB}=\frac{8}{10}=\frac{4}{5}[/tex].

Solution:
The ratio of the perimeters equals the similarity ratio: [tex]k=\frac{60}{36}=\frac{5}{3}[/tex]. The third side of the first triangle is [tex]36-9-15=12[/tex].
So the sides of the second triangle are [tex]9 \times \frac{5}{3}=15;\ 12 \times \frac{5}{3}=20;\ 15 \times \frac{5}{3}=25[/tex].

Solution:
The ratio of the areas of similar triangles equals the square of the similarity ratio: [tex]\frac{S_1}{S_2}=\left(\frac{6}{5}\right)^2=\frac{36}{25}[/tex].
So [tex]S_2=108 \times \frac{25}{36}=75[/tex].

Solution:
Angles with respectively parallel sides are either equal or supplementary. If two pairs of corresponding angles were supplementary, these four angles would add up to a full angle, which is impossible, because two angles of a triangle add up to less than a straight angle.
So at least two pairs of angles are equal, e.g. [tex]\angle A=\angle A_1[/tex] and [tex]\angle B=\angle B_1[/tex], and by the AA criterion [tex]\triangle ABC \sim \triangle A_1B_1C_1[/tex].

Solution:
The right triangles share the angle at [tex]B[/tex], so [tex]\triangle CBD \sim \triangle ABC[/tex], hence [tex]\frac{BD}{CB}=\frac{CB}{AB}[/tex].
So [tex]BD=\frac{CB^2}{AB}=\frac{144}{18}=8[/tex] and [tex]AD=18-8=10[/tex].

Solution:
Since [tex]AB \parallel CD[/tex], [tex]\triangle EDC \sim \triangle EAB[/tex], so [tex]\frac{ED}{EA}=\frac{EC}{EB}=\frac{DC}{AB}=\frac{5}{13}[/tex].
[tex]\frac{ED}{ED+8}=\frac{5}{13}[/tex] gives [tex]ED=5[/tex], and [tex]\frac{EC}{EC+10}=\frac{5}{13}[/tex] gives [tex]EC=\frac{25}{4}[/tex].

Solution:
The upper side of the square cuts off a triangle similar to the given one, with base [tex]x[/tex] and height [tex]9-x[/tex], so [tex]\frac{x}{18}=\frac{9-x}{9}[/tex], i.e. [tex]9x=162-18x[/tex] and [tex]x=6[/tex].

Solution:
Let [tex]M[/tex] be the midpoint of [tex]AB[/tex], [tex]CH[/tex] the altitude and [tex]K[/tex] the foot of the perpendicular from [tex]G[/tex] to [tex]AB[/tex]. The centroid lies on the median [tex]CM[/tex] and [tex]MG=\frac{1}{3}MC[/tex]. Since [tex]GK \parallel CH[/tex], the right triangles with vertex [tex]M[/tex] are similar, so [tex]\frac{GK}{CH}=\frac{MG}{MC}=\frac{1}{3}[/tex].
Hence [tex]GK=\frac{12}{3}=4[/tex].

Solution:
Since [tex]PQ \parallel AC[/tex], [tex]\triangle PBQ \sim \triangle ABC[/tex] with ratio [tex]\frac{BQ}{BC}=\frac{3}{5}[/tex]. The perimeters of similar triangles are in the same ratio, so [tex]P_{PBQ}=\frac{3}{5} \times 35=21[/tex].

Solution:
Twice the area equals each side times its altitude: [tex]a \times h_a=b \times h_b=2S[/tex], so [tex]\frac{a}{b}=\frac{h_b}{h_a}=\frac{5}{3}[/tex].
Hence [tex]a=\frac{5}{8} \times 40=25[/tex].

Solution:
The sides of [tex]\triangle A_1B_1C_1[/tex] are midsegments of [tex]\triangle ABC[/tex]: [tex]B_1C_1=\frac{BC}{2};\ C_1A_1=\frac{CA}{2};\ A_1B_1=\frac{AB}{2}[/tex]. So [tex]\triangle A_1B_1C_1 \sim \triangle ABC[/tex] with ratio [tex]k=\frac{1}{2}[/tex], and [tex]S_{A_1B_1C_1}=\frac{1}{4} \times 120=30[/tex].

Solution:
The ratio of the areas is the square of the similarity ratio: [tex]\frac{S_1}{S_2}=\left(\frac{4}{5}\right)^2=\frac{16}{25}[/tex]. Let [tex]S_1=16t;\ S_2=25t[/tex]; then [tex]41t=82[/tex], [tex]t=2[/tex], and [tex]S_1=32;\ S_2=50[/tex].

Solution:
The centroid divides the median in the ratio [tex]2:1[/tex] from the vertex, so [tex]CG=\frac{2}{3}CM[/tex]; also [tex]CP=\frac{1}{2}CM[/tex].
Hence [tex]PG=CG-CP=\frac{1}{6}CM[/tex], so [tex]CM=30[/tex].

Solution:
Let [tex]M[/tex] be the midpoint of the hypotenuse. The median to the hypotenuse equals half of it: [tex]CM=\frac{AB}{2}=9[/tex]. The centroid divides the median in the ratio [tex]2:1[/tex] from the vertex, so [tex]GC=\frac{2}{3}CM=6[/tex].

Solution:
By the angle bisector theorem [tex]\frac{AB}{BC}=\frac{AL}{LC}=2[/tex]. In the right triangle [tex]\cos\angle ABC=\frac{BC}{AB}=\frac{1}{2}[/tex], so [tex]\angle ABC=60^\circ[/tex].

Solution:
Let [tex]K[/tex] and [tex]H[/tex] be the feet of the perpendiculars from [tex]D[/tex] and [tex]B[/tex] to [tex]AC[/tex]. Then [tex]DK \parallel BH[/tex], so [tex]\triangle ADK \sim \triangle ABH[/tex] and [tex]\frac{DK}{BH}=\frac{AD}{AB}=\frac{8}{12}=\frac{2}{3}[/tex].

Solution:
Let the midsegments parallel to the corresponding sides be such that [tex]m_1=n_1;\ m_2=n_2;\ m_3=n_3[/tex]. Each midsegment is half of the side it is parallel to, so [tex]BC=2m_1=2n_1=B_1C_1[/tex] and similarly [tex]CA=C_1A_1;\ AB=A_1B_1[/tex].
By the SSS criterion [tex]\triangle ABC \cong \triangle A_1B_1C_1[/tex].