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Geometry Level 3

Let $\triangle ABC$ rectangle in $\angle B$ and $AB<BC$ . Take a point $P$ in bisector of $\angle ABC$ such that $AP$ is perpendicular to this bisector. Let $M$ the middle point of $AC$ , and $E$ the interseccion of $MP$ with $AB$ . If $EM=15$ find $BC$

The answer is 30.

1 solution

Paola Ramírez
Jan 28, 2015

Let $F$ the point where $AP$ cuts $BC$ .

Then $\triangle ABP$ and $\triangle ABP$ and $\triangle BPF$ . So $AP=BP=FP \Rightarrow P$ i middle point of $AF$ . By Thales' theorem we have that $MP||FC \Rightarrow EM||BC$ and as $M$ is middle point, $E$ is middle point of $AB$

$\therefore BC=2EM=\boxed{30}$

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The answer is 30.

1 solution

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