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.

**Next chapter :**** In isosceles triangle, lines joining from vertex to base are equal in length**