# Congruent Triangles Problems by Side-Angle-Side

Problem 1
Let ABC be an isosceles triangle with AC=BC. Let D denote the midpoint of AB(i.e., D is the point on AB for which AD=BD). Prove that triangle ACD is congruent to triangle BCD
Solution:
We are going to prove that the two triangles are congruent by Side-Side-Side. From the condition we have AD = BD (D is the midpoint); AC = BC (triangle ABC isosceles); CD is in both triangles. So triangles ACD and BCD have two congruent sides and the included angle. Therefore triangles ACD and BCD are congruent.

Problem 2
Let ABC be an isosceles triangle. Prove that the altitude drawn to the base is the median and the angle bisector.
Solution:
From the previous question, we have CD as the bisector of the base AB in the isosceles triangle ABC. We proved that triangles ACD and BCD are congruent. From the congruence it follows that angle ADC = CD. Also notice that their sum is 180°, therefore angle ADC = CDB = 90°, which shows that CD is the altitude. Also from the congruence of the two triangles we know that CAD = BAD, i.e. CD is the bisector of angle A.

Problem 3
Let ABC be an isosceles triangle with AC=BC. Prove that the angles BAC and ABC are equal to each other.
Solution:
According to Problem 1, the triangles ACD and BCD are congruent. The angles BAC and ABC are corresponding angles in congruent triangles, therefore they are equal to each other.

Problem 4
Calculate the perimeter of an isosceles triangle ABC if the perimeter of the triangle ADC is 18 cm., CD = 6 cm. and AD = BD (fig.5)

Solution:
The perimeter of the triangle ADC = AC + CD + AD = 18 <=> AC + 6 + AD = 18 <=> AC + AD = 12
Notice AC = BC (the triangle is isosceles) and AD = DB follows that AC + AD = DB + BC = 12
Therefore 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:

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

Problem 6
If the intersection of the diagonals of a parallelogram is O, then prove that the opposite sides are equal to each other.
Solution:

ABCD is a parallelogram. Therefore AO = OC and BO = OD. So triangles AOD and BOC are congruent (side-angle-side, AO = OC; BO = OD and angles DOA = BOC – apex) therefore AD = BC Triangles AOB and DOC are also congruent and therefore AB = CD.

Problem 7
Let's use a triangle ABC (AC < BC) - with squares BMNC and CPQA built outwardly from this triangle. Prove that AN = BP

Solution:
According to the picture in the triangles BPC and ACN BC = CN (BMNC – square);
PC = AC (CPQA - square). Also angle BCP = ACN = 90° + ACB. Therefore triangles are congruent and therefore AN = BP

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