Question

in an isosceles triangle ABC with AB=AC ,BD and CE are two medians. prove that BD=CE.

Answer 1

In triangle CDB and BEC,
BC=BC [common side]
angle B = angle C [given thet ABC is isoceless]
AC = AB => DC = EB [AC and AB are medians]
Thus, triangle CDB congruent to triangle BEC.
Therefore, BD=CE. {corresponding parts of congruent triangles}


Hope it helps you. Thank you.

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! )