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(H): In △ ABC...

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(H): In }△ ABC,D is mid point of }\overline{BC} . }E is foot of perpendicular } from }C on external angle bisector of }\angle BAC . prove 2 DE = } AB+ AC

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△

∠

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A B C, ∠ B A C = 90^{\circ},

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⊥ A C

D E \times (A B + A C) = A B \times A C

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

∠ B A C = 90^{0}

⊥

D E \times (A B + A C) = A B \times A C

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