Let $ABCD$ be a parallelogram. Show that $\angle BQD = 90^ \circ$.

Question

Let $ABCD$ be a parallelogram. There is point $P$ on the $BD$ such that $AP=BD$. $Q$ is the midpoint of the $CP$. Show that $\angle BQD = 90^ \circ$.

Answer 1

Alternatively, draw the second diagonal and denote the center with $O$.

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Using the formulas of medians: $$\begin{align}DQ^2&=\frac{2CD^2+2PD^2-PC^2}{4};\\ BQ^2&=\frac{2BC^2+2BP^2-PC^2}{4};\\ OP^2&=\frac{2AP^2+2PC^2-AC^2}{4} \Rightarrow \color{red}{AC^2}=2AP^2+2PC^2-4OP^2=\color{red}{2BD^2+2PC^2-4OP^2}.\end{align}$$ Adding: $$\begin{align}DQ^2+BQ^2&=\frac{(2CD^2+2BC^2)+2(PD^2+BP^2)-2PC^2}{4}=\\ &=\frac{(\color{red}{AC^2}+BD^2)+2\left(\left(\frac{BD}{2}+OP\right)^2+\left(\frac{BD}{2}-OP\right)^2\right)-2PC^2}{4}=\\ &=\frac{(\color{red}{2BD^2+2PC^2-4OP^2}+BD^2)+2\left(\frac{BD^2}{2}+2OP^2\right)-2PC^2}{4}=\\ &=BD^2.\end{align}$$

Answer 2

Translate $P$ by $\vec{AD}$ into new point $R$. Then triangle $BDR$ is isosceles and $BCRP$ is a parallelogram. Since $Q$ is a midpoint of a diagonal of parallelogram $BCRP$ it halves $BR$ also. But then $DQ$ is an altitude in isosceles triangle $BDR$ and thus a conclusion.