prove that $AQ^2+CP^2=AC^2+PQ^2$

Question

In a Right triangle $ABC$ right angled at $B$, $P$ and $Q$ are points on sides $AB$ and $AC$ respectively. EDIT: Actually options were given in my book as follows in which option $(c)$ was given correct.

$(a)$$AQ^2+CP^2=2(AC^2+PQ^2)$

$(b)$ $2(AQ^2+CP^2)=AC^2+PQ^2$

$(c)$$ AQ^2+CP^2=AC^2+PQ^2$

$(d)$ $AQ+CP=\frac{1}{2}(AC+PQ)$

I have drawn a perpendicular $QM$ on to $AB$.

Applying Pythagoras theorem for $\Delta BPC$ we get

$$CP^2=BP^2+BC^2$$ but $$BC^2=AC^2-AB^2$$ so

$$CP^2=BP^2-AB^2+AC^2 \tag{1}$$

Applying Pythagoras theorem for $\Delta AQM$ we get

$$AQ^2=QM^2+AM^2 \tag{2}$$

adding $(1)$ and $(2)$ we have

$$AQ^2+CP^2=BP^2-AB^2+AC^2+QM^2+AM^2$$

but i am stuck here

Answer 1

It seems to me that more assumptions are required.

For instance, if you take an isosceles right triangle with sides $2\sqrt{2}$, $2\sqrt{2}$ and $4$, $P$ and $Q$ midpoints of $\overline{AB}$ and $\overline{AC}$, respectively, as in the picture below you have \begin{align} |AQ|^2+|CP|^2&=4+10=14&&\text{while}&|AC|^2+|PQ|^2&=16+2=18 \end{align}

Answer 2

I don't think you've given all the information or the statement is false. Your final expression seems correct and working from there we basically want to prove: $$ AQ^2+CP^2=BP^2-AB^2+AC^2+QM^2+AM^2 = AC^2 + PQ^2 $$ This reduces to: $$PQ^2 = BP^2 -AB^2 +QM^2 +AM^2 $$ The angle at $M$ is right. $$ PQ^2 = MP^2 + QM^2$$ Plugging these equations together yields $$MP^2 = BP^2 + AM ^2 -AB^2 \ (1)$$ $MP^2 $ is obviously a positive number but $ AB^2 = (AM + MP + BP)^2 $ so of course $AB^2 > BP^2 + AM^2$ so the righthandside of equation $(1)$ is negative. So what we want to prove is equal to proving something impossible. So the question must be wrong.