Question

Prove that the angles opposite to equal sides of a triangle are equal.

Answer 1

Hint: To solve this question, we will use the special properties of the isosceles triangle that two of its sides are equal and that the altitude is the same as the median. Further, in this problem, we will use the properties of congruence to prove that the angles opposite to equal sides of a triangle are equal.

Complete step-by-step answer:
Before solving the problem, we should know the basics of how we can apply the results of congruency to get the answer to the above problem in hand. For reference to the above solution, we will refer to the following diagram of triangle ABC below.

Thus, we have a triangle ABC.
Now we will construct a line CD which is perpendicular to the side AB.
We are given that two sides are equal (in this case, we take AC=BC). We would then have to prove that angle A = angle B. Thus, we will consider two triangles, triangle ADC and triangle BDC. From both these triangles, we have,
AC = BC (according to question)
Angle ADC = angle BDC = 90 degrees (CD is the altitude a shown)
CD = CD (common side of both triangles)
Thus, according to the RHS criteria of congruence, which states that two triangles are congruent if both are right-angled, hypotenuse are equal (AC = BC) and one other side is equal (CD = CD). Thus triangles ADC and BDC are congruent to each other. This means all their properties and values are exactly the same. Thus, we can conclude that angle A = angle B.
Hence, proved.

Note: In this problem, we used the RHS criteria to prove the congruence. We also have many other congruence criteria like SSS (side, side side), SAS (side, angle and side), ASA (angle, side and angle) and AAS (angle, angle and side). For example, if two triangles have all the sides of the same length, we can use the SSS criteria to prove their congruency.