In this chapter we will try to prove that in isosceles triangle, two points on base that are equidistant from the extremities of are also equidistant from the opposite vertex.
Given:
Consider the given isosceles triangle ABC in which AB = AC.
Point M & N are points on base of isosceles triangle such that BM = CN
To prove;
The point M and N are equidistant from vertex A.
i.e. AM = AN
Proof
Consider the triangle AMB and ANC
AB = AC { equal sides of isosceles triangle }
BM = NC { given }
∠ABM = ∠ ACN { angle opposite to equal sides are equal }
By SAS congruency condition, triangle AMB and ANC are congruent.
i.e. \mathtt{\triangle AMB\ \cong \triangle ANC}
Since both the triangles are congruent, we can say that AM = AN.
Hence, proved.
Next chapter : In equilateral triangle, all angles are equal