Congruent Triangles Problems by Angle-Side-Angle and Side-Side-Side

Angle-Side-Angle

If a side and two angles of a triangle are equal to side and two angles of another triangle, the two triangles are congruent.

The following property is determined:

If in one triangle two of the sides are equal the opposite angles are also equal.

Side-Side-Side:

If the sides of a triangle are congruent to the sides of another triangle, the two triangles are congruent.

Problem 1
Prove that the bisectors to the thighs of the isosceles triangle are equal.

Solution:
Lets take ABC as isosceles with base AB and AM and BN are the bisectors of angle CAB and angle CBA (fig.38) Triangles AMB and BNA are congruent (angle-side-angle) because:
1. angle CAB = angle CBA
2. AB – in both triangles.
3. angle MAB = angle NBA = ¹/₂ angle CAB Segments AM and BN are corresponding in these congruent triangles and therefore AM = BN

Problem 2
Prove the perpendiculars built from the ends of a side to the median, built to that side, are equal.

Solution:
Lets mark the observed triangle with ABC and CM is a median to AB.(fig.39)
From A and B we built the perpendiculars AA₁ and BB₁ to the straight line CM .Triangles AA₁M and BB₁M are congruent (angle-side-angle) because angle BB₁M = angle AA₁M = 90°, angle AMA₁ = angle BMA₁ opposite, AM = BM. Then AA₁ = BB₁ as corresponding sides in these triangles.

Problem 3
Prove that the perpendiculars built from any point of the bisector of an angle towards his shoulders cut from them equal segments.

Solution:
Lets take that for angle AOB point M is unspecified point from his bisector OL.(fig.40)
Lets take that MP perpendiculiar OA and MQ OB To prove that OP = OQ is enough to prove that triangle OPM Congruent Triangles OQM . But OPM OQM(angle-side-angle), because OM = OM, angle QOM = angle POM (OL is bisector), angle OQM = angle OPM = 90°, from where OP = O

Problem 4
Prove that if in a triangle the altitude and the bisector are built from the same apex the triangle is isosceles.

Solution:
Lets take that for triangle ABC the altitude and the bisector from apex C match. (fig.41) To prove that AC = BC, i.e. ABC is isosceles it is enough to prove that ADC and BDC are congruent . But ADC Congruent Triangles BDC (angle-side-angle) because angle ACD ≤ angle BCD (CD is bisector) angle ACD ≤ angle CDB = 90° ( CD is ), CD = CD, from where
AC = BC

Problem 5
Prove that two triangles are congruent if they have corresponding :
à) side, included angle and bisector of the angle
b) two angles and an altitude
c) two angles and bisector
d) two sides and a median to the third side

Solution:
a) Lets take that for triangles ABC and A₁B₁C₁ the following elements are equal: AC = A₁C₁, angle A = angle A₁, AL = A₁L₁ are bisectors of angles A and A₁(fig.42) We look at ALC and A₁L₁C₁. They are congruent (side-angle-side), because AC = A₁C₁ (by condition), AL = A₁L₁(by condition), angle CAL = angle C₁A₁L₁ (half of equal angles), therefore angle C = angle C₁ Then triangle ABC Congruent Triangles A₁B₁C₁(angle-side-angle)

b) If the two triangles have two equal angles the third angles are also equal(the sum of the three angles of every triangle is 180°) Then there is no difference if the equal altitudes are built from the given apexes or from the third apex. Lets take that for triangles ABC and A₁B₁C₁ (fig.43) are given angle A ≤ angle A₁ angle C ≤ angle C₁ and the altitudes BH and B₁H₁ also equal. Then ABH Congruent Triangles A₁B₁H₁(angle-side-angle) because BH = B₁H₁
angle H ≤ angle H₁ = 90° ABC and A₁B₁C₁ (fig.43) are given angle A ≤ angle A₁ angle C ≤ angle C₁ and the altitudes BH and B₁H₁ also equal. Then ABH A₁B₁H₁(angle-side-angle) because BH = B₁H₁ angle H ≤ angle H₁ = 90° angle A = angle A₁ From the congruence of these two triangles follows that AB = A₁B₁. Therefore ABC Congruent Triangles A₁B₁C₁(angle-side-angle) because AB = A₁B₁, angle A ≤ angle A₁ è angle C ≤ angle C₁

c) As in b) here there is no difference if the bisectors are built from the given angles or from the third. Lets take that for triangles ABC and A₁B₁C₁ we know that angle A = angle A₁ angle ACL = angle A₁C₁L₁(halves of equal angles), CL = C₁L₁. From the congruence of these two triangles follows that AC = A₁C₁ and because angle A = angle A₁, angle C = angle C₁ we get ABC Congruent Triangles A₁B₁C₁(angle-side-angle)

d) Lets take that for triangles ABC and A₁B₁C₁ we know that AB = A₁B₁; AC = A₁C₁; AM = A₁M₁, where AM and A₁M₁ are medians of BC and B₁C₁(fig.45) On rays AM and A₁M₁ we take points D and D₁ so that AM = MD = A₁M₁ = M₁D₁
Then AMC Congruent Triangles BM D (side-angle-side) because AM = MD, BM = CM(AM is a median), angle AMC= angle BMD (apex angles).
Analogically and A₁M₁C₁ B₁M₁D₁. From the congruence of the triangles follows that AC = BD and A₁C₁ = B₁D₁, so AC = BD = A₁C₁ = B₁D₁
We look at ABD è A₁B₁D₁ By condition AB = A₁B₁, BD = B₁D₁ (from what we examined so far and from AD = 2AM, A₁D₁ = 2 A₁M₁ follows that AD = A₁D₁ Then ABD Congruent Triangles A₁B₁D₁ (side-side-side)
Analogically we prove that ADC A₁D₁C₁ From the last two congruence of triangles follows that angle DAB = angle D₁A₁B₁ and angle DAC = angle D₁A₁C₁ Then angle CAB = angle C₁A₁B₁ (sum of equal angles). By side-angle-side ABC Congruent Triangles A₁B₁C₁ çàùîòî AB = A₁B₁, AC = A₁C₁ and angle CAB = angle C₁A₁B₁

Congruent triangles problems by Side-Angle-Side

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