Question : In a triangle, $ABC, A P$, the bisector of $\angle A$, is perpendicular to $B C$ at point $P$. The measures of $BP$ and $PC$ are $x$ and $3y$, respectively. The measures of $AB$ and $AC$ are $4 x$ and $(5 y+21)$, what is the value of $(x+y)$?
Option 1: 21
Option 2: 15
Option 3: 12
Option 4: 18
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Correct Answer: 12
Solution :
In $\triangle APB$ and $\triangle APC$,
$\angle APB = \angle APC=90°$
$\angle BAP = \angle CAP$
AP = AP (common side)
$⇒\triangle APB$ ~ $\triangle APC$
$⇒\frac{PB}{PC} = \frac{AB}{AC}$
$⇒\frac{x}{3y} = \frac{4x}{5y+21}$
$⇒12y = 5y+21$
$⇒y = 3$ --------------(i)
$AP$, the bisector of $\angle A$, is perpendicular to $BC$.
⇒ AP is the perpendicular bisector of BC.
i.e., BP = PC
$\therefore x = 3y =3×3 = 9$
So, $x+y = 3+9 = 12$
Hence, the correct answer is 12.
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