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

Angle-Side-Angle

If in one triangle 2 sides and 2 included angles are the same measure as two sides and two included angles from the second triangle, these triangles also equal.

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 CAB and CBA (fig.38) Triangles AMB and BNA are congruent (angle-side-angle) because:
1. CAB = CBA
2. AB – in both triangles.
3. MAB = NBA = 1/2 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 BB1M = AA1M = 90°, AMA1 = 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 AOB point M is unspecified point from his bisector OL.(fig.40)
Lets take that MP OA and MQ OB To prove that OP = OQ is enough to prove that OPM OQM . But OPM OQM(angle-side-angle), because OM = OM, QOM = POM (OL is bisector), OQM = 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 are equal, the triangle is isosceles.

Solution:
Lets take that for 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 BDC (angle-side-angle) because ACD ≤ BCD (CD is bisector) ACD ≤ 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, A = 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), CAL = C1A1L1 (half of equal angles), therefore C = C1 Then ABC 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 A ≤ A1 C ≤ C1 and the altitudes BH and B1H1 also equal. Then ABH A1B1H1(angle-side-angle) because BH = B1H1
H ≤ H1 = 90° ABC and A1B1C1 (fig.43) are given A ≤ A1 C ≤ C1 and the altitudes BH and B1H1 also equal. Then ABH A1B1H1(angle-side-angle) because BH = B1H1 H ≤ H1 = 90° A = A1 From the congruence of these two triangles follows that AB = A1B1. Therefore ABC A1B1C1(angle-side-angle) because AB = A1B1, A ≤ A1 and C ≤ 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 A = A1 ACL = A1C1L1(halves of equal angles), CL = C1L1. From the congruence of these two triangles follows that AC = A1C1 and because A = A1, C = C1 we get ABC 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 BM D (side-angle-side) because AM = MD, BM = CM(AM is a median), AMC= 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 A1B1D1 (side-side-side)
Analogically we prove that ADC A1D1C1 From the last two congruence of triangles follows that DAB = D1A1B1 and DAC = D1A1C1 Then CAB = C1A1B1 (sum of equal angles). By side-angle-side ABC A1B1C1 because AB = A1B1, AC = A1C1 and CAB = C1A1B1

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