In figure, ∠BAC=90o, AD is its bisector. If DE ⊥ AC, Prove that DE

1.	In figure, ∠BAC=90o, AD is its bisector. If DE ⊥ AC, Prove that DE×(AB+AC)=AB×AC.
Answer» In figure, $∠ B A C = 90^{o}$ , AD is its bisector. If DE $⊥$ AC, Prove that $D E \times (A B + A C) = A B \times A C$ .

In figure, ∠BAC=90o, AD is its bisector. If DE ⊥ AC, Prove that DE×(AB+AC)=AB×AC.

Answer»

In figure, $∠ B A C = 90^{o}$ , AD is its bisector. If DE $⊥$ AC, Prove that $D E \times (A B + A C) = A B \times A C$ .