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RHS Criteria for Congruency

19. In ΔABC, ...

Question

19. In ΔABC, Angle BAC = 90^°, AD is the bisector of angle BAC and DE is perpendicular to AC. Prove that - DE (AB+AC) = AB AC.

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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

A B C, ∠ B A C = 90^{\circ},

A D

D E

⊥ A C

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

∠ B A C = 90^{o}

D E ⊥ A C

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

Δ

∠

= 90^{o}

⊥

\times

+

=

\times

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