# Congruent Triangles Problems by Side-Angle-Side

Problem 1
Triangle ABC – isosceles, CD is bisector to the base AB. Prove that ACD is congruent to BCD
Solution:
Prove that the two triangles are congruent by Side-Angle-Side. From the condition we have angle ACD = DCB (CD is bisector); AC = BC (triangle ABC isosceles); CD is in both triangles. So triangles ACD and BCD have congruent two sides and the included angle. Therefore triangle ACD and BCD are congruent.

Problem 2
Prove that the altitude, the bisector and the median to the base of a isosceles triangle match.
Solution:
From the previous problem we have the CD is bisector in the isosceles triangle ABC and we proved that triangles ACD and BCD are congruent. From the congruence follows that angle ADC = CDB but they are contiguous therefore their sum is 180°, from where angle ADC = CDB = 90°, which show that CD is an altitude. From the congruence of the two triangle we have that AD = BD, i.e. CD is a median.

Problem 3
Prove that the angles in the base of every isosceles triangle are equal.
Solution:
We use again problem 1 and that triangles ACD and BCD are congruent. The angles in the base of isosceles triangle are equal because they are corresponding angles in congruent triangles.

Problem 4
Calculate the perimeter of isosceles triangle ABC if the perimeter of >ADC is 18 cm., and CD = 6 cm. and AD = BD (fig.5)
Solution:
The perimeter of triangle ADC = AC + CD + AD = 18 <=> AC + 6 + AD = 18 <=> AC + AD = 12
Because AC = BC (the triangle is isosceles) and AD = DB follows that AC + AD = DB +BC = 12
The perimeter of triangle ABC = AB + AC + BC = AD + DB + AC + BC = 12 + 12 = 24cm.

Problem 5
Prove that a straight line which cuts equal segments from the shoulders of an angle is perpendicular to the bisector of that angle.
Solution:

Lets take that the straight line cuts from the shoulders of angle AOB equal segments OC = OD. Then triangle OCD is isosceles and OF is bisector to the base. According to problem 2 OF is an altitude, i.e. OF is perpendicular to α

Problem 6
Prove that if the diagonals of a quadrangle halve each other every two opposite sides are equal.
Solution:

For the quadrangle ABCD we know that AO = OC è BO = OD. Then triangles AOD and BOC are congruent (side-angle-side, AO = OC; BO = OD and angles DOA = BOC – apex) therefore AD = BC Analogically triangles AOB and DOC are congruent from where AB = CD.

Problem 7
Prove that if triangle ABC is congruent to A1B1C1 and for points M and M1 from sides AB and A1B1 it is in force that AM = A1M1 CM = C1M1 and angle BMC = B1M1C1
Solution:

Because triangle ABC is congruent to triangle A1B1C1 AB = A1B1, BC = B1C1, CA = C1A1 and angles A = A1, B = B1, C = C1. Then triangle AMC is congruent to triangle A1M1C1 (side-angle-side), from where CM = C1M1. MB = M1B1 because from the equal segments AB and A1B1 we subtract the equal segments AM and A1M1. Angles BMC and B1M1C1 are equal because they are equal contiguous angles to angles AMC and A1M1C1, which are equal (triangle AMC is congruent to A1M1C1)

Problem 8
We have triangle ABC (AC < BC). Outwardly are built:
squares BMNC and CPQA. Prove that AN = BP

Solution:
For triangles BPC and ACN BC = CN (BMNC – square);
PC = AC (CPQA - square) angle BCP = ACN = 90° + ACB. Therefore triangles are congruent from where AN = BP

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