Question

In figure BD = AB and BD² = BEX BC. Prove that AD bisects CAE.​

Answer 1

Answer:

n ΔABC

AD is the bisector of ∠BAC

∠1=∠2 …(i)

Also in ΔADC

∠3=∠2+∠C (ext. angle is equal to sum of opposite interior angles)

∠3>∠2 (ext. angle is greater than one of the interior angles)

But ∠1=∠2

∠3>∠1

AB>BD

Hence proved