# Line joining vertex base to midpoint of other side in isosceles triangle

In this chapter we will prove the concept that the line joining the vertex of base to mid point of the opposite side of the isosceles triangle are equal.

Given:
Given above is the isosceles triangle in which AB = AC and BC is the base.

Line BM is drawn from base vertex B and meet at mid point of opposite side AC such that AM = MC.

Similarly, line CN is drawn from base vertex C and meet at midpoint of opposite side AB such that AN = NB.

To prove:
Prove that BM = CN

Proof:
Since ABC is an isosceles triangle, two sides are equal.
AB = AC

The expression can also be written as;

\mathtt{\frac{1}{2} AB\ =\frac{1}{2} \ AC}\\\ \\ \mathtt{BN=\ CM}

Taking triangle BNC and BMC
NB = MC { Proved above }
∠NBC = ∠MCB { angle opposite to equal sides are equal }
BC = CB { Common side }

By SAS congruency condition, both triangle BNC and BMC are congruent.
i.e. \mathtt{\triangle BNC\ \cong \triangle BMC}

Since both the triangles are congruent, we can say that BM = CN

Hence in isosceles triangle, line drawn from base to midpoint of opposite sides are equal.