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The Mid-Point Theorem

Question 6 In...

Question

Question 6
In the given figure, AC = AE, AB = AD and $∠ B A D = ∠ E A C .$ Show that BC = DE.

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Solution

$I t i s g i v e n t h a t ∠ B A D = ∠ E A C ∠ B A D + ∠ D A C = ∠ E A C + ∠ D A C [A d d i n g ∠ D A C o n b o t h s i d e s] ∠ B A C = ∠ D A E, I n Δ B A C a n d Δ D A E, A B = A D (G i v e n) ∠ B A C = ∠ D A E (P r o v e d a b o v e) A C = A E (G i v e n) ∴ Δ B A C ≅ Δ D A E (B y S A S c o n g r u e n c e r u l e) ∴ B C = D E (B y C P C T)$

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The Mid-Point Theorem

Standard IX Mathematics

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