In $\triangle ABC$, $AB=AC$, $I$ is the incenter, and $PD^2=PE\cdot PF$. Show that $\square IPBC$ is a cyclic quadrilateral.

Question

In $\triangle ABC$, $AB=AC$, $I$ is the incenter, and $PD^2=PE\cdot PF$. Show that $\square IPBC$ is a cyclic quadrilateral.

Answer 1

Hint: The weird condition that $PD^2=PE\cdot PF$ is, since the triangle is isosceles, equivalent to $\square PFCD \sim \square PDBE$. Can you continue from there to show that $\angle BPC = \angle BIC$?

Answer 2

Let $\measuredangle BAC=\alpha$.

Since $\measuredangle EPD=\measuredangle DPF$ and $\frac{PD}{PE}=\frac{PF}{PD}$, we obtain $\Delta EPD\sim\Delta DPF$,

which gives $\measuredangle DEP=\measuredangle FDP$ and $\measuredangle EDP=\measuredangle DFP$ and since $BEPD$ and $CFPD$ are cyclics,

we obtain: $$\measuredangle BPC=180^{\circ}-(\measuredangle PBD+\measuredangle PCD)=180^{\circ}-(\measuredangle PED+\measuredangle PCD)=$$ $$=180^{\circ}-(\measuredangle PDF+\measuredangle PCD)=180^{\circ}-(\measuredangle PCF+\measuredangle PCD)=$$ $$=180^{\circ}-\measuredangle ACB=180^{\circ}-\left(90^{\circ}-\frac{\alpha}{2}\right)=90^{\circ}+\frac{\alpha}{2}.$$ In another hand, $$\measuredangle BIC=180^{\circ}-\frac{1}{2}(\measuredangle ABC+\measuredangle ACB)=180^{\circ}-\frac{1}{2}(180^{\circ}-\alpha)=90^{\circ}+\frac{\alpha}{2}.$$ Id est, $\measuredangle BPC=\measuredangle BIC$, which gives that $B$, $I$, $P$ and $C$ are cocyclics.

Done!

Answer 3

HINT (A way to solve).-For IPBC to be a cyclic quadrilateral it is necessary and sufficient that the circumcircle to any trio of points of the four I, P, B, C, coincide. You have the formulas you need in Wikipedia if you do not know them.

1) Let $$A=(0,b),B=(-a,0) , C=(a,0),\overline{BC}=2a, \overline{AB}=\overline{AC}=\sqrt{a^2+b^2}$$

2) Incenter $I=\left(0,\dfrac{ab}{a+\sqrt{a^2+b^2}}\right)$

3) Circumcircle of triangle $\triangle {BCI}$: an equation is obtained making the determinant equal to $0$ of the matrix $$\begin{pmatrix} x^2+y^2&x&y&1\\a^2&-a&0&1\\a^2&a&0&1\\\dfrac{a^2b^2}{(a+\sqrt{a^2+b^2})^2}&0&\dfrac{ab}{a+\sqrt{a^2+b^2}}&1\end{pmatrix}$$ 4) You have to verify that the point $P$ is in this circle.