Please solve q 85 ?

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Answer 1 · 2018-05-20T02:02:16

9

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For the triangles $Delta PAC$ and $Delta PAB$ to be similar, all three angles must be the same.

It is easy to see that $/_BPA$ is common
$/_ACP$ is obtuse - so it can not be equal to $/_ABP$ , and thus it must be equal to the remaining angle, $/_PAB$
Finally, we must have $/_PAC=/_ABP$

By the property of similar triangles we have

$AC : AP:PC = AB:BP:AP$

Denoting $PC$ and $AP$ by $x$ and $y$ , respectively we have (using $BP = BC+PC = 7+x$ )

$6/8 = y/(7+x) = x/y$

Hence $y^2 = x(7+x)$ and we have

$6^2/8^2 = x^2/y^2 = x/(7+x)$

and thus

$x/(7+x) = 9/16 qquad implies qquad x/7 = 9/(16-9) = 9/7$

Thus
$x = 9$