Given,
In isosceles Δ ABC,
BD and CE are bisectors of ∠B and ∠C
And,
AB = AC
To prove: BD = CE
Proof:
In Δ BEC and Δ CDB, we have
∠B =∠C (Angles opposite to equal sides)
BC = BC (Common)
∠BCE = ∠CBD (Since, ∠C = ∠B\(\frac{1}{2}\)∠C = \(\frac{1}{2}\)∠B ∠BCE = ∠CBD)
By ASA theorem, we have
Δ BEC ≅ Δ CDB
EC = BD (By c.p.c.t)
Hence, proved