In the given figure, BD bisects $∠ A B C$ and BD is perpendicular to AC. Find x + y.

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Solution

In $Δ A B D$ and $Δ C B D$

(i) $∠ A B D = ∠ C B D$ ... (given )

(ii) $B D = B D$ ……(common)

(iii) $∠ B D A = ∠ B D C = 90^{\circ}$ ….(given)

$∴ Δ A B D ≅ Δ C B D$ (ASA postulate)

$\Rightarrow A D = D C$ ……..(CPCT)
$5 x - 3 = 2 x + 6 3 x = 9 x = 3$

and $B A = B C$ ……..(CPCT)
$2 y - 1 = 13 2 y = 14 y = 7$
x+y = 3 + 7 = 10

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