# Diagonals of the rectangle formed by the angle bisectors of a parallelogram

Let $$ABCD$$ be a parallelogram. I proved that the angle bisectors of $$A$$, $$B$$, $$C$$, $$D$$ form a rectangle. How can I prove that the diagonals of this rectangle are parallel to the sides of $$ABCD$$? And is there a relation between the length of these diagonals and $$AB$$ or $$BC$$?

I'm looking for an elementary solution only using parallelograms, congruent triangles. • Do you know proportional segment theorem (Thales)? The diagonals are going to be equal to sides of the parallelogram. – Vasya Apr 5 at 16:59
• sorry, I thought you meant midpoints. – Vasya Apr 5 at 18:16
• That is to prove that EG parallel to AD and find a relation between EG and the sides AD or AB. – user661240 Apr 5 at 19:12

Let the angle bisectors at $$A$$ and $$B$$ meet at $$P$$. Drop perpendiculars from $$P$$ to points $$A^\prime$$, $$B^\prime$$, $$Q$$ on the sides of the parallelogram as shown: Clearly, we have constructed a few similar right triangles, and, in particular, two pairs of congruent right triangles. We see that $$P$$ must be halfway between opposite sides of the parallelogram; likewise for $$P^\prime$$. This guarantees the desired parallelism property. $$\square$$

For a relation about the lengths, lop-off the trapezoid on one side and paste it to the other, getting a rectangle whose width is equal to the original base of the parallelogram, $$\overline{AD}$$. Then, for the configuration shown (where $$|AD|>|AB|$$):

$$|AD| = a + b + d = |AB| + d \qquad\to\qquad d = |AD| - |AB|$$

• Awesome!....[+1] – Dr. Mathva Apr 5 at 20:03 Without loss of generality, we can assume that $$AB>BC$$. Let's start with proving that $$AW=ZB=DU=VC=CD-AD$$.

In parallelogram, adjacent angles are supplementary so $$m\angle{A}+m\angle{D}=180\circ$$. From $$\triangle{ADV}$$ we have: $$0.5 \cdot m\angle{A}+m\angle{D}+m\angle{DVA}=180\circ$$ which means $$m\angle{DVA}=0.5 \cdot m\angle{A}$$. Thus, $$\triangle{ADV}$$ is isosceles and $$DV=AD$$, $$VC=CD-AD$$.

Similarly, $$CU=BC$$ and $$DU=CD-BC=CD-AD=VC$$.

It's easy to show now that $$\triangle{AZM} \cong \triangle{DMV}$$: $$AZ=AD=DV$$ and alternative interior angles are congruent. Thus, $$MZ=MD$$. Similarly, $$\triangle{AMD} \cong \triangle{BCS}$$ so $$BS=MD=MZ$$. Because $$DZ||BU$$, $$MSBZ$$ is a parallelogram and $$MS||AB||DC$$.

We can now see that $$MS=CD-AD=TN$$ (diagonals of rectangle are congruent).

• Is this exercise too hard for middle school or am i too bad in geometry? BTW AMD not AZD is congruent to BCS. Thank you. – user661240 Apr 5 at 19:26
• @user661240: It's pretty tough for middle school level. Thanks for your correction. – Vasya Apr 5 at 19:57