Question

$∆$ ABC is isosceles in which AB = AC. Seg BD and seg CE are medians. Show that BD = CE.

Answer 1

​In △ABC, AB = AC
∴ ∠B = ∠C (Angles opposite to equal sides)
Dividing both sides by 2, we get
$\frac{1}{2}\angle \mathrm{B}=\frac{{\displaystyle 1}}{{\displaystyle 2}}\angle \mathrm{C}$
⇒ ∠EBC = ∠DCB ...(2)
In △BEC and △CDB
∠EBC = ∠DCB [From (2)]
BE = CD (E and D are the mid points of AB & AC respectively and AB = AC)
BC = CB (Common)
By SAS test of congruency
△BEC ≅ △CDB
∴ BD = CE (corresponding sides of congruent triangles)