euclidean-geometrygeometry
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.
Hope the following sketch can help.
Since $DM\perp AC$ and $M$ is a midpoint of $AC$, we obtain $AD=DC$.
Can you end it now?
Since $DMNC$ is an isosceles trapezoid, we obtain: $$\measuredangle PMN=\measuredangle PDC=\measuredangle PCD=\measuredangle PNM,$$ which gives $$PM=PN.$$ Also, $$\Delta MDN\cong\Delta NCM,$$ which gives $$\measuredangle HMN=\measuredangle HNM,$$ which gives $$HM=HN,$$ which says that $PMHN$ is a kite, which says $$PH\perp MN.$$ About a kite see here: https://en.wikipedia.org/wiki/Kite_(geometry)
Best Answer
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.