In the given figure, BD bisects $∠ A B C$ and BD is perpendicular to AC. Find x + y.
10

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Solution

The correct option is A 10
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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In the given figure, BD bisects ∠ABC and BD is perpendicular to AC. Find x + y.10

The correct option is A 10In ΔABD and ΔCBD (i) ∠ABD=∠CBD ... (given ) (ii) BD=BD……(common) (iii) ∠BDA=∠BDC=90∘ ….(given) ∴ΔABD≅ΔCBD (ASA postulate) ⇒AD=DC ……..(CPCT) 5x–3=2x+63x=9x=3 and BA=BC ……..(CPCT) 2y–1=132y=14y=7 x+y = 3 + 7 = 10

In the given figure, BD bisects $∠ A B C$ and BD is perpendicular to AC. Find x + y.
10