Suppose ABC is an isosceles triangle with AB=AC ; BD and CE are bisectors of ∠B and ∠C. Prove that BD=CE
In △ABC
AB=AC [Given]
∠ABC=∠ACB......(1) [Angles opposite to equal sides of a trinagle are equal]
Also, 12AB=12AC
BE=CD.......(2) [Halves of equals are equal]
Since BD and CE are two medians
Now,
In △BDC and △CEB
From (1) ∠BCD=∠CBE
From (2) BE=CD
BC=CB [Common]
∴△BDC≅△CEB [SAS Congruence rule]
∴BD=CE [By CPCT]