$ABCD$ is a convex quadrilateral in which $AB^2+CD^2=BC^2+AD^2$. Prove by vectors that diagonals $AC$ and $BD$ are orthogonal to each other.

Question

$ABCD$ is a convex quadrilateral in which $AB^2+CD^2=BC^2+AD^2$. Prove by vectors that diagonals $AC$ and $BD$ are orthogonal to each other.

How should i prove it? I think Ptolemy's theorem is not working here.

Please help me.

Answer 1

Let me try. Let $O$ be an intersection of $AC$ and $BD$

Vector method:

One has $$\begin{align} \vec{AO} + \vec{OB} = \vec{AB} \implies AO^2 + OB^2 + 2\vec{AO}\cdot\vec{OB} &= AB^2.\\ BO^2 + OC^2 + 2\vec{BO}\cdot\vec{OC} &= BC^2.\\ CO^2 + OD^2 + 2\vec{CO}\cdot\vec{OD} &= CD^2.\\ DO^2 + OA^2 + 2\vec{DO}\cdot\vec{OA} &= DA^2. \end{align}$$

From $AB^2+CD^2= BC^2+DA^2$, we have $$\begin{align} \vec{AO}\cdot\vec{OB} + \vec{CO}\cdot\vec{OD} &= \vec{BO}\cdot\vec{OC} + \vec{DO}\cdot\vec{OA} \\ \vec{AO}\cdot\vec{OB} + \vec{CO}\cdot\vec{OD} + \vec{OB}\cdot\vec{OC} + \vec{OD}\cdot\vec{OA} &= \vec{0} \\ \vec{AC}\cdot\vec{OB} + \vec{CA}\cdot\vec{OD} &= \vec{0} \\ \vec{AC}\cdot\vec{DB} &= \vec{0} \end{align}$$ Then $AC \perp BD$.

Answer 2

Since $$\vec{AC}=\vec{AB}+\vec{BC}=\vec{AD}+\vec{DC}$$ we have $$\vec{AB}-\vec{DC}=\vec{AD}-\vec{BC}.$$ Square both sides and use $AB^2+CD^2=AD^2+BC^2$, we get $$\vec{AB}\cdot\vec{DC}=\vec{AD}\cdot\vec{BC}.$$ That means $$(\vec{AB}+\vec{BC})\cdot\vec{DC}=(\vec{AD}+\vec{DC})\cdot\vec{BC},$$ or $$\vec{AC}\cdot \vec{DC}=\vec{AC}\cdot\vec{BC}.$$ So $\vec{AC}\cdot\vec{BD}=0$.