Hello, muks!

1. $\displaystyle P$ is a point on side $\displaystyle BC$ of a square $\displaystyle ABCD.$

From point $\displaystyle B$, a perpendicular is drawn to $\displaystyle AP$, cutting $\displaystyle DC$ at $\displaystyle Q.$

Prove: .$\displaystyle AP=BQ$ Let $\displaystyle BQ$ intersect $\displaystyle AP$ at $\displaystyle R.$

Code:

Q
D * - - - - * - - - - * C
| * |
| * |
| * |
| * |
| * *P
| R* β|
| * * |
| * *α|
| * α β *|
A * - - - - - - - - - * B

In right triangle $\displaystyle ARB$, let $\displaystyle \angle RAB = \alpha$

Then $\displaystyle \angle RBA = \beta$, the complement of $\displaystyle \alpha$

Hence, $\displaystyle \angle RBP = \alpha$, the complement of $\displaystyle \beta$

Right triangles $\displaystyle PBA\text{ and }QCB$ have: .$\displaystyle \begin{Bmatrix}\text{an equal side:} & AB \:=\:BC \\ \text{an equal angle: } & \angle PAB \:=\:\angle QBC \end{Bmatrix} $

Therefore: .$\displaystyle \Delta PBA \cong \Delta QCB \quad\Rightarrow\quad AP \:=\:BQ$