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**