In the given figure, BD bisects ∠ABC and BD is perpendicular to AC. Find, x + y.

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Solution

The correct option is C 10
In $△ A B D$ and $△ C B D$ ,
(i) $∠ A B D = ∠ C B D$ --- (given)
(ii) $B D = B D$ --- (Common)
(iii) $∠ B D A = ∠ B D C = 90^{\circ}$ --- (given)

$∴ △ A B D ≅ △ C B D$ (ASA postulation)

$\Rightarrow A D = D C$ --- (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.

The correct option is C 10In △ABD and △CBD, (i) ∠ABD=∠CBD --- (given) (ii) BD=BD --- (Common) (iii) ∠BDA=∠BDC=90∘ --- (given) ∴△ABD≅△CBD (ASA postulation) ⇒AD=DC --- (CPCT) 5x−3=2x+6 3x=9 x=3 and BA=BC --- (CPCT) 2y−1=13 2y=14 y=7 x+y=3+7=10