$PA^2 + PB^2 + PC^2 + PD^2 = 4\cdot R^2$

Question

Let $AB$ and $CD$ be two chords of a circle having radius $R$, which intersect at the point $P$ in right angle.

Then prove that $PA^2 + PB^2 + PC^2 + PD^2 = 4\cdot R^2$.

Please anyone help. I am stuck here.

Answer 1

Hint: $$PA^2+PC^2=AC^2=(2R\sin\angle PBC)^2$$ $$PB^2+PD^2=BD^2=(2R\sin\angle PCB)^2$$

where I used the law of sines in the last equality.

You can see that: $$\sin\angle PBC=\cos\angle PCB$$ because they are complementary angles.

Alternative solution

We can see that:

$$PA^2+PC^2+PB^2+PD^2=AC^2+BD^2$$

Now, if we consider the right triangles $\triangle ACE$ and $\triangle APC$, we get:

$$AC:d=AP:AD$$

because they are similar; where $d$ is the diameter $AE$. Hence $AC={AP\over AD}d$.

We can get an analogous relation for $BD$, considering the similar right triangles $\triangle BDF$ and $\triangle BPC$:

$$BD:d=BP:BC$$

so $BD={BP\over BC}d$.

Replacing above these relations, we get:

$$PA^2+PC^2+PB^2+PD^2=AC^2+BD^2=\left[\left({AP\over AD}\right)^2+\left({BP\over BC}\right)^2\right]d^2$$

Now we can consider the similar right triangles $\triangle APD$ and $\triangle BPC$:

${BP\over BC}={PD\over AD}$

hence: $\left[\left({AP\over AD}\right)^2+\left({BP\over BC}\right)^2\right]d^2=\left[\left({AP\over AD}\right)^2+\left({PD\over AD}\right)^2\right]d^2=d^2=4R^2$