Question

in an isosceles triangle abc with ab =ac bd and ce are two medians prove That bd = ce

Answer 1

we can prove it by congruency as
AE=BE(as CE is median)
AD=CD(as BD is median)
Therefore,
in triangle BFC and triangle CDB
BF=CD(proved above) 
angle DBC= angle ECB(it is an isocsles triangle and median eill be at same lenghts)
BC=BC(common)

triangle BFC= triangle CDB (by SAS congruence rule)

bd=ce(CPCT)
Hence Proved

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!