Got stumped on another problem when I was tutoring a student in Geometry. Here it is:
Given: $ABCD$ is a parallelogram,
$\overline{AM},\overline{BN}$ angle bisectors,
$DM=4\,\text{ft.}$, $MN=3\,\text{ft.}$
Find: the perimeter of $ABCD$
I saw there is a theorem that the angle bisectors of a parallelogram create a rectangle, so in other words, if we construct angle bisectors $\overline{DO}$ and $\overline{CP}$ such that $O,P\in\overline{AB}$, then $MNOP$ is a rectangle, correct? (EDIT: I'm wrong. See below.) The key seems to be finding $CN$ and $AD$ — then we can find the other sides and the perimeter since the opposite sides are congruent. How should we go about this from here? Keep in mind, the student does not know trig, but they do know the Pythagorean Theorem. Thank you!
EDIT: Actually, the theorem would apply like this: Let $E$ be the intersection of the extensions of $\overline{AM}$ and $\overline{BN}$, let $\overline{DG}$ and $\overline{CG}$ be angle bisectors meeting at the point $G$, let $F=\overline{BN}\cap\overline{CG}$, and let $H=\overline{AM}\cap\overline{DG}$. Then $EFGH$ is a rectangle. Could this help in this problem?
Best Answer
We have two isosceles triangles that we can use.
As $AB \parallel DC$, $\angle BAM = \angle AMD = \angle DAM$. So, $DA = DM = 4$.
Similarly $NC = CB = DA = DM$
$DC = DM + MN + NC = 2 DM + MN = 11$
Perimeter of the parallelogram is $2 (DC + DA) = 30$