Isosceles triangle two points on base equidistance from vertex

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.

Property of isosceles triangle

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

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

Leave a Comment

Your email address will not be published. Required fields are marked *

You cannot copy content of this page