Question

ABC is an isosceles triangle with AB =

AC and BD and CE are its two medians.

Show that BD = CE.
​

Answer 1

Answer:

Given:-

AB = AC

Also , BD and CE are two medians

Hence ,

E is the midpoint of AB and

D is the midpoint of CE

Hence ,

1/2 AB = 1/2AC

BE = CD

In Δ BEC and ΔCDB ,

BE = CD [ Given ]

∠EBC = ∠DCB [ Angles opposite to equal sides AB and AC ]

BC = CB [ Common ]

Hence ,

Δ BEC ≅ ΔCDB [ SAS ]

BD = CE (by CPCT)

Answer 2

Answer:

A simpler way:

In ∆ABC,

AB = AC (given)

=>

In ∆EBC and ∆DCB,

BC = BC (common side)

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! )