In $\Delta ABC , D$ and $E$ are midpoints of the sides $AB$ and $AC$ respectively, $CD$ and $BE$ intersect at $P$ with $\angle BPC = 90^\circ$.

Question

In $\Delta ABC , D$ and $E$ are midpoints of the sides $AB$ and $AC$ respectively, $CD$ and $BE$ intersect at $P$ with $\angle BPC = 90^\circ$. Suppose $BD = 8\sqrt{5} , CE = 6\sqrt{5}$ . Find $BC$ .

What I Tried: Here is a picture :-

It was not long that I made some wonderful observations. For example, I see that $P$ is the median of this triangle. So probably I can assume $DP = x$ and $PE = y$ , and $PC = 2x$ and $PB = 2y$ , and they would work fine as we got a lot of right angles too?

Also from Midpoint Theorem, $DE$ is half of $BC$. So finding $DE$ alone, will be enough, but I have not worked out a way for that.

Another thing I thought of is let $[\Delta DBP] = M$ , $[\Delta PBC] = N$ , $[\Delta PCE] = O$ , $[DPEA] = K$ . Then I get :-
$$\rightarrow M + N = O + K$$
$$\rightarrow N + O = M + K$$

I think I have a lot of information, but I cannot combine all of these together (or maybe some of them), to get a good result. I need a bit of help with that.

Can anyone help me? Thank You.

Answer 1

Hint:By Pythagoras in $\Delta PDB,\Delta PEC$ $$x^2+4y^2=320$$,$$4x^2+y^2=180$$