In a trapezoid with perpendicular diagonals the square of sum of the bases is equal to sum of the squares of the diagonals

Question

Diagonals of a trapezoid are perpendicular to each other. Prove that the square of sum of the bases of the trapezoid is equal to sum of the squares of the diagonals.

One way to solve this problem is to extend $AB$ from $B$ and then from $C$ drawing a line parallel to $BD$. Assume the intersection is $F$. Now it can be proven that the quadrilateral $BFCD$ is parallelogram hence $\angle ACF= 90^{\circ}$. And after applying the Pythagorean theorem, the problem is solved. But I'm trying to solve this problem with a different approach, Here we want to prove $$(a+b)^2=(n+q)^2+(m+p)^2$$ $$a^2+b^2+2ab= n^2+q^2+2nq+m^2+p^2+2mp$$ Since $a^2+b^2 = m^2+n^2+p^2+q^2$ the problem is reduced to proving $ab= mp+nq$. Here I'm not sure how to continue. Since $\triangle ABO$ and $\triangle CDO$ are similar triangles we have $\frac ab=\frac mp = \frac nq$ but don't know if this helps.

Answer 1

You have stopped at $$ \frac ab=\frac mp=\frac nq. $$

Multiply now both sides of the equation $b^2=p^2+q^2$ by $\displaystyle\frac ab$ to obtain $$ ab=\frac ab(p^2+q^2)=\frac mpp^2+\frac nqq^2=mp+nq. $$

Answer 2

Let $\frac{a}{b}=k \implies a=kb, m=kp, n=kq$

Then we have $ab=kb^2, mp+nq=kp^2+kq^2=k(p^2+q^2)=kb^2$.

Answer 3

This is probably not the approach you had in mind, but I am fond of vector methods in proving geometric statements.

Write $\overrightarrow{AB} = \vec x$, $\overrightarrow{AD} = \vec y$, and note that $\overrightarrow{DC} = \lambda\vec x$ for some scalar $\lambda$ because the quadrilateral is a trapezoid. Let $\overrightarrow{AC}=\vec u = \vec y+\lambda\vec x$ and $\overrightarrow{BD}=\vec v = \vec y - \vec x$ be the diagonals; then we are given $\vec u\cdot \vec v = 0$.

It follows that $\|\vec u\|^2 + \|\vec v\|^2=\|\vec u-\vec v\|^2$. But $\vec u-\vec v = (1+\lambda)\vec x$, and so the sum of the squares of the lengths of the diagonals equals $\|(1+\lambda)\vec x\|^2 = (\|\vec x\|+\|\lambda\vec x\|)^2$, as desired.