# Angle bisectors of exterior angles in trapezoid

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 $$MK=15$$ $$cm.$$

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.

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.

• 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? Sep 2, 2019 at 11:31
• @AndrewRogers If P is the center, then PM = PD and therefore $\angle PMD = \angle PDM = 0.5(ext. \angle D)$. Then $\angle MPD = 180^0 - 0.5(ext. \angle D) - 0.5(ext. \angle D)$.
– Mick
Sep 2, 2019 at 12:51

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 $$N$$.

• Thank you! I appreciate it! Sep 2, 2019 at 14:34

Thank you! And let me finish the problem.

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

$$MK=15$$ $$cm$$

$$MK=MP+PN+NK$$

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

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