# Prove equality between segments of a circumference

Let $A,B,C,D$ be points in a circumference, such that $\overline {AB}$ is a diameter and the chords $\overline {AC}$ and $\overline {BD}$ intersect in a point $P$ inside the circumference. Prove that $AB^2$ = $AP \cdot AC$ $+$ $BP \cdot BD$

(I made a image of what the problem means if i wrote something wrong)

My try

I tried a lot of things and i can't get to the answer, i tried Ptolemy's theorem with the cyclic quadrilateral $ABCD$ but i don't see a way to prove the statement.

Any hints?

Let Q is a point in AB such that $PQ \perp AB$. One then has $$AP.AC=AQ.AB$$ $$BP.BD = BQ.BA$$

Thus, one has $AP.AC + BP.BD = AB^2$.

• Thanks, very easy – Rodrigo Pizarro Feb 7 '18 at 1:59

Let $X$ be on $AB$ such that $PX\perp AB$. Since angles $ADB$ and $BCA$ are right angles we have $$\triangle AXP\sim \triangle ACB\quad\hbox{and}\quad \triangle BXP\sim\triangle BDA\ ,$$ so $$\frac{BX}{BP}=\frac{BD}{AB}\quad\hbox{and}\quad \frac{AX}{AP}=\frac{AC}{AB}$$ and $$AB=AX+XB=\frac{AP\cdot AC}{AB}+\frac{BP\cdot BD}{AB}\ .$$

Hint: by the power of point, $PA\cdot PC = PB \cdot PD = PO^2 - R^2\,$, where $O$ is the center of the circle and $R = AB/2$ its radius, so: $$AP(AP+PC)+BP(BP+PD)=AP^2+BP^2+2(PO^2-R^2)$$

Now consider that $PO$ is a median in $\triangle ABP\,$, and reduce the RHS to $AB^2\,$.

\begin{align*} \frac{AP}{AB}\cdot\frac{AC}{AB}+\frac{BP}{AB}\cdot\frac{BD}{AB}&=\frac{\sin\angle ABP}{\sin\angle APB}\cdot\frac{\sin\angle ABC}{\sin\angle ACB}+\frac{\sin\angle BAP}{\sin\angle APB}\cdot\frac{\sin\angle BAD}{\sin\angle ADB}\\ &=\frac{\sin\angle ABP\sin\angle ABC+\sin\angle BAP\sin\angle BAD}{\sin\angle APB\sin\frac{\pi}{2}}\\ &=\frac{\sin\angle ABP\cos\angle BAP+\sin\angle BAP\sin\angle ABP}{\sin\angle APB}\\ &=\frac{\sin(\angle ABP+\angle BAP)}{\sin\angle APB}\\ &=\frac{\sin(\pi-\angle APB)}{\sin\angle APB}\\ &=1 \end{align*}

A different method. Note that $$\angle ACB$$ is a right angle; also by the crossed-chords theorem, $$AP\cdot PC=BP\cdot PD$$. Therefore \eqalign{AB^2 &=AC^2+BC^2\cr &=(AP+PC)^2+BP^2-PC^2\cr &=AP^2+AP\cdot PC+BP\cdot PD+BP^2\cr &=AP\cdot AC+BP\cdot BD\ .\cr}