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