Thread: Arc over right angle proof

1. Arc over right angle proof

On two sides of right angle with vertex in point O we choose points A and B, |OA|=|OB|. On arc AB with center in O we choose point P. Through that point we draw line parallel with AB which intersects OA in point C, and OB in point D. Prove that |PC|2+|PD|2=|AB|2.

I hope you understand my question. I find a lot of similar and right triangles but always get stuck.

2. Re: Arc over right angle proof

Hello, kicma!

On two sides of right angle $O$ we choose points A and B, |OA|=|OB|.
On arc AB with center in O we choose point P.
Through that point we draw line parallel with AB
which intersects OA in point C, and OB in point D.
Prove that: |PC|2+|PD|2=|AB|2.

Code:
      |
D *
| *
|   *
B *     *
| *     *  P
|   *     o
|     * .   *
r |     . *     *
|   .     *     *
| . θ   45o *  45o *

- + - - - - - - * - - -* - - -
|O     r      A      C
We have quarter circle $AOB.$
. . $r = AO = OB,\;\angle BAO = \angle ABO = 45^o.$
Note that: $AB \,=\, \sqrt{2}r \quad\Rightarrow\quad \overline{AB}^2 \,=\,2r^2$

$P$ is on arc $AB$.
$CPD \parallel AB.\;\;\angle DCO = \angle CDO = 45^o.$
Let $\theta = \angle POC.$

We will use the Law of Sines.

$\text{In }\Delta OPC\!:\;\frac{PC}{\sin\theta} = \frac{r}{\sin45^o} \quad\Rightarrow\quad PC \,=\,\frac{r\sin\theta}{\sin45^o}$

$\text{In }\Delta DOP\!:\;\frac{PD}{\sin(90-\theta)} = \frac{r}{\sin45^o} \quad\Rightarrow\quad PD \,=\,\frac{r\cos\theta}{\sin45^o}$

$\text{Hence: }\:\overline{PC}^2 + \overline{PD}^2 \;=\;\frac{r^2\sin^2\theta}{\frac{1}{2}} + \frac{r^2\cos^2\theta}{\frac{1}{2}}$

. . . . . $\overline{PC}^2 + \overline{PD}^2 \;=\;2r^2\sin^2\theta + 2r^2\cos^2\theta$

. . . . . $\overline{PC}^2 + \overline{PD}^2 \;=\;2r^2(\sin^2\theta + \cos^2\theta)$

Therefore: . $\overline{PC}^2 + \overline{PD}^2 \;=\;2r^2\;=\;\overline{AB}^2$

3. Re: Arc over right angle proof

I forgot to mention I'm not allowed to use Law of Sines/Cosines. There should be a way to prove this without it if you can help.