Question

in an isosceles triangle ABC With AB=AC BD and CE are two median prove that BD=CE

Answer 1

Hope this answer will be helpful

Answer 2

Answer:

A simpler way:

In ∆ABC,

AB = AC (given)

=> <ABC = <ACD (opposite sides are equal)

In ∆EBC and ∆DCB,

BC = BC (common side)

<EBC = <DCB ( <ABC = <ACB)

BE = DC (E & D are mid points on AB & AC)

=> ∆EBC is congruent to ∆DCB (by SAS

criteria)

=> BD = CE ( by c.p.c.t ) ( proved! )