# Thread: diagonals of a rectangle and a square

1. ## diagonals of a rectangle and a square

If we have a square ABCD and a rectangle AEFD. Let P be the intersection of AC and ED, Q the intersection of AF and BD. How can we prove that the line PQ is parallel to AD?

2. Hello, geo_math!

We have a square $\displaystyle ABCD$ and a rectangle $\displaystyle AEFD$.
Let $\displaystyle P$ be the intersection of $\displaystyle AC$ and $\displaystyle ED$, $\displaystyle Q$ the intersection of $\displaystyle AF$ and $\displaystyle BD.$

Prove that $\displaystyle PQ$ is parallel to $\displaystyle AD.$
Code:
    Q *
: * *
:   *   *
:     *     *
:       *       *
:         *      θ  *   A
:         B * - - - - - * - - - - - - - - - - - * E
:           | *     α * |   *  θ         θ  *   |
:           |   *   * α |       *       *       |
:           |     *     |           *           |
:           |   * R * α |       *       *       |
:           | *     α * |   *  θ         θ  *   |
:         C * - - - - - * - - - - - - - - - - - * F
:         *      θ  *  D
:       *       *
:     *     *
:   *   *
: * *
P *

Let $\displaystyle AC$ and $\displaystyle BD$ intersect at $\displaystyle R.$
Note that all angles labeled $\displaystyle \theta$ are equal.
. . And all angles labeled $\displaystyle \alpha$ are 45°.
Further note that: .$\displaystyle AC \perp BD\:\text{ and }\:RA = RD$

In right triangles $\displaystyle PRD$ and $\displaystyle QRA:\!\;\;\angle PDR \,=\,\angle QAR \,=\,\theta + 45^o,\;\;RD = RA$

Hence: .$\displaystyle \Delta PRD \cong \Delta QRA \quad\Rightarrow\quad PR \,=\,QR$

Then $\displaystyle \Delta PRQ$ is an isosceles right triangle: .$\displaystyle \angle PQR \,=\,45^o$

Since $\displaystyle \angle QDA = 45^o,\;\;PQ \parallel AD\quad\text{(alternate-interior angles)}$