Given: AB = BC and D is the midpoint of AC
The above triangle is an Isoscele triangle with two sides equal.
Angles opposite to the equal sides of an isosceles triangle are also equal.
Proof: Consider an isosceles triangle ABC where AB = BC. We need to prove that the angles opposite to the sides AB and BC are equal, that is, ∠BAD = ∠BCD. ...
Sides opposite to the equal angles of a triangle are equal.
BD = BD (Common side)
∠ABD = ∠BDC = 90° (By construction)
Thus, ∆BAD ≅ ∆BCD (congruence criterion)
so AB = BC
Hence prooved
< A = < C