# Please solve q 85 ?

May 20, 2018

9

#### Explanation:

For the triangles $\Delta P A C$ and $\Delta P A B$ to be similar, all three angles must be the same.

• It is easy to see that $\angle B P A$ is common
• $\angle A C P$ is obtuse - so it can not be equal to $\angle A B P$, and thus it must be equal to the remaining angle, $\angle P A B$
• Finally, we must have $\angle P A C = \angle A B P$

By the property of similar triangles we have

$A C : A P : P C = A B : B P : A P$

Denoting $P C$ and $A P$ by $x$ and $y$, respectively we have (using $B P = B C + P C = 7 + x$)

$\frac{6}{8} = \frac{y}{7 + x} = \frac{x}{y}$

Hence ${y}^{2} = x \left(7 + x\right)$ and we have

${6}^{2} / {8}^{2} = {x}^{2} / {y}^{2} = \frac{x}{7 + x}$

and thus

$\frac{x}{7 + x} = \frac{9}{16} q \quad \implies q \quad \frac{x}{7} = \frac{9}{16 - 9} = \frac{9}{7}$

Thus
$x = 9$