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In a trapezoid $ABCD$ angle bisectors of exterior angles at apices $A$ and $D$ intersect in $M$ and angle bisectors of exterior angles at apices $B$ and $C$ intersect in $K$. Find $P_{ABCD}$ if $MK=15$ $cm.$

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So it's easy to see $\triangle AMD$ and $\triangle BKC$ are right triangles. I noticed $MK$ is the midsegment of $ABCD$ but I don't know how to show $P$ and $N$ are midpoints of, respectively, $AD$ and $BC$. Would appreciate help of any kind.

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3 Answers 3

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Let $P$ and $N$ be the midpoints of $AD$ and $BC$ respectively. Then, $PN||AB||CD$. This means $\angle DPN = ext. \angle D$ by alternate angles.

Note that $P$ is also the center of the circle passing through $A, M, D$. Then, $\angle MPD = 180^\circ – ext. \angle D$ by angle sum of triangle.

This means $MPN$ is a straight line. Likewise, $KNP$ is also a straight line.

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  • $\begingroup$ Thank you for your response! I am not sure I understood why $\angle MPD = 180^\circ - ext \angle D$. Can you explain this to me? $\endgroup$ Sep 2, 2019 at 11:31
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    $\begingroup$ @AndrewRogers If P is the center, then PM = PD and therefore $\angle PMD = \angle PDM = 0.5(ext. \angle D)$. Then $\angle MPD = 180^0 - 0.5(ext. \angle D) - 0.5(ext. \angle D)$. $\endgroup$
    – Mick
    Sep 2, 2019 at 12:51
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Let the intersection of bisector at $B$ and the extension of $DC$ be $E$ and the intersection of bisector at $A$ and the extension of $DC$ be $F$. $\triangle ADF$ and $\triangle BCE$ are isosceles and we have:

$MF=MA$, that is $M$ is the midpoint of $AF$.

$KE=KB$, that is $K$ is the midpoint of $BE$.

Therefore $MK||AB$ so it divides $AD$ and $BC$ at their midpoints $P$ and $N$.

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  • $\begingroup$ Thank you! I appreciate it! $\endgroup$ Sep 2, 2019 at 14:34
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Thank you! And let me finish the problem.

$P_{ABCD} = AB + BC + CD + AD$

$MK=15$ $cm$

$MK=MP+PN+NK$

$15=\dfrac{1}{2}AD + \dfrac{AB+CD}{2} + \dfrac{1}{2}BC$ $/.2$

$30=AB+BC+CD+AD$.

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