Prove that the sides opposite the congruent base angles of a triangle are congruent.

The sides opposite to congruent base angles are congruent.

Step-by-step explanation:

Let AB be the base, with C denoting the top of the triangle ABC. We are given that <BAC is congruent to <ABC.

Let D be a point on AB such that the line segment CD is perpendicular to the base (in other words, CD is the height of the triangle), thus creating two triangles: ACD and BCD, that share a common side CD.

Since <ADC = <BDC = 90 degrees, it follows that <ACD is congruent to <BCD (using the fact that knowing two of the angles are congruent between two triangles, so must be their third angles).

We use the angle-side-angle (ASA) postulate now: Since the side CD is shared, i.e, congruent, by triangles ACD and BCD, and two enclosing angles <ACD=<BCD and <ADC=<BDC are also congruent, it follows that the triangles ACD and BCD are congruent. This implies the sides AC and CB, which are the sides opposite of the congruent base angles of ABC, are congruent - which was to be proven.

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