Question

ABC is an isosceles triangle in which AB = AC and BD and CE are its two medians. show that BD = CE

Answer 1

ar(ABC)= 1/2×AC×BD ----(1)
ar(ABC)= 1/2×AB×CE
= 1/2×AC×CE (AB=AC,GIVEN) ----(2)

From (1) & (2)
1/2×AC×BD = 1/2×AC×BD
Therefore AC = BD

Answer 2

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! )