If $A, B, P$, and $Q$ are four harmonic points and $O$ and $M$ are the midpoints of $AB$ and $PQ$ respectively, then $(OB)^2 + (OP)^2 = (OM)^2$

Question

I already computed that $OB^2 = (OP)(OQ)$ and $MP^2 = (MA)(MB)$, I tried to sum both equations but I can't get the result.

Here is the same question in another post: Prove that $OP^2=PC^2+OB^2$. but I want to see if it's possible to show it with the equations that I got.

Answer 1

I think the correct formula is $(OB)^2 + (MP)^2 = (OM)^2$?

Set a coordinate system on a line: $O(0)$, $A(2a)$, $B(-2a)$, $P(2p)$ and $Q(2q)$. Then $M(p+q)$. Now you have to prove $$4a^2+(p-q)^2=(p+q)^2 $$ which is the same as $a^2=pq$. Now, since $${AP\over PB}:{AQ\over QB}=-1$$ we get $${a-p\over a+p}={q-a\over a+q}\implies a^2=pq$$ and we are done.