[asy][asy] /* File unicodetex not found. */ /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(11cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = 2, xmax = 7.5, ymin = -3.5, ymax = 3; /* image dimensions */ /* draw figures */ draw(circle((4.28,-1.12), 1.75)); draw((3.28,2.8)--(7.04,-1.7)); draw((7.04,-1.7)--(4.06,-3.18)); draw((4.06,-3.18)--(2.36,-1.82)); draw((2.36,-1.82)--(3.28,2.8)); draw((3.28,2.8)--(4.06,-3.18)); draw((4.28,-1.12)--(3.8,-1.18)); draw((3.8,-1.18)--(7.04,-1.7)); draw((3.8,-1.18)--(2.36,-1.82)); /* dots and labels */ dot((5.62,0),dotstyle); label("$E$", (5.66,0.06), NE * labelscalefactor); dot((3.28,2.8),dotstyle); label("$A$", (3.32,2.85), NE * labelscalefactor); dot((7.04,-1.7),dotstyle); label("$B$", (7.07,-1.65), NE * labelscalefactor); dot((2.57,-0.78),dotstyle); label("$H$", (2.6,-0.72), NE * labelscalefactor); dot((2.36,-1.82),dotstyle); label("$D$", (2.39,-1.77), NE * labelscalefactor); dot((4.28,-1.12),dotstyle); label("$O$", (4.32,-1.07), NE * labelscalefactor); dot((3.19,-2.48),dotstyle); label("$G$", (3.22,-2.43), NE * labelscalefactor); dot((4.06,-3.18),dotstyle); label("$C$", (4.09,-3.12), NE * labelscalefactor); dot((5.05,-2.69),dotstyle); label("$F$", (5.09,-2.63), NE * labelscalefactor); dot((3.8,-1.18),dotstyle); label("$P$", (3.83,-1.13), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Let $O$ be the center of the inscribed circle and let $E$, $F$, $G$, and $H$ be the points of tangency of the incircle to sides $AB$, $BC$, $CD$, and $DA$, respectively. Since $\angle OPC = \angle OFC = 90^\circ$, pentagon $OPGCF$ is cyclic with diameter $OC$. Similarly, pentagon $OPHAE$ is also cyclic with diameter $OA$. By the Law of Sines, \[\frac{\sin \angle EPB}{\sin \angle PEB} = \frac{BE}{BP} = \frac{BF}{BP} = \frac{\sin \angle FPB}{\sin \angle PFB}\] In addition, we have \begin{align*}\frac{\sin \angle EPB}{\sin \angle FPB} = \frac{\sin \angle PEB}{\sin \angle PFB} = \frac{\sin (180^\circ - \angle PEA)}{\sin (180^\circ - \angle PFC)} = \frac{\sin (180^\circ - \angle POA)}{\sin (180^\circ - \angle POC)} = \frac{\sin \angle POA}{\sin \angle POC}\end{align*} and \[\frac{\sin \angle POA}{\sin \angle POC} = \frac{\sin (180^\circ - \angle AHP)}{\sin (180^\circ - \angle PGC)} = \frac{\sin \angle PHD}{\sin \angle PGD} = \frac{\sin \angle HPD}{\sin \angle GPD}\] Thus $\frac{\sin \angle HPD}{\sin \angle GPD} = \frac{\sin \angle EPB}{\sin \angle FPB}$, and so we have that $\angle HPD = \angle EPB$ and $\angle GPD = \angle FPB$. Therefore, \begin{align*} \angle APB &= \angle APE + \angle BPE \\&= \angle AHE + \angle DPH \\&= \angle AEH + \angle DPH \\&= \angle APH + \angle DPH \\&= \angle APD\end{align*} as desired.

by Newton theorem, $EH$,$FG$,$BD$ are concurrent at $M$. Apply Brianchon theorem to $AHDCFB$, $AEBCGD$ and $AC$,$BD$,$EG$,$HF$ are concurrent at $J$. Apply Pascal theorem to $EEHGGF$, $HHEFFG$ and $K$,$M$,$L$ are collinear. $AC$ is the polar of $M$ and $IM$ is perpendicular to $AC$ and $I$,$P$,$M$ are collinear. $1=(KB,KC/KJ,KL)=(B,D/J,M)$ and $\angle JPM=90^{\circ}$. Therefore $\angle APB=\angle APD$

Let $R,S,T,U$ be the points of tangency of the incircle, and let $Q=RU \cap ST$, and $L,M$ be the midpoints of $RS,TU$. Since $\triangle QRS \sim \triangle QTU, \angle QLR =\angle QMT$, which means that $\angle QLI +\angle QMI =180$. But $Q,L,M$ are the inverses of $P,B,D$, so this means that $\angle BPI +\angle DPI=180$, which means $\angle BPC=\angle DPC$

robinpark wrote: Thus $\frac{\sin \angle HPD}{\sin \angle GPD} = \frac{\sin \angle EPB}{\sin \angle FPB}$, and so we have that $\angle HPD = \angle EPB$ and $\angle GPD = \angle FPB$. Why is this true? Thanks in advance!

First we apply an inversion with center $O$ and power the radius of inscribed circle. We get the following problem

: Let $EFGH$ be a cyclic quadrilateral $A^{'},B^{'},C^{'},D^{'}$ midpoint of the sides $EH,EF,FG,GH$.Let $EH$,$FG$ intersect in $P^{'}$.Prove that $\angle OB^{'}P^{'}+\angle OD^{'}P^{'}=180^{\circ}$ For this we employ complex numbers (as ususal we denote the coordinate of a point by its small case letter)

Bump ! Can someone answer this ? MathPanda1 wrote: Post #2 wrote: Thus $\frac{\sin \angle HPD}{\sin \angle GPD} = \frac{\sin \angle EPB}{\sin \angle FPB}$, and so we have that $\angle HPD = \angle EPB$ and $\angle GPD = \angle FPB$. Why is this true? Thanks in advance!

solution becomes pretty easy if we use inversion with center O, since A’, B’, C’ and D’ become midpoints of line segments H’E’, E’F’, F’G’, G’H’, where E, F, G, H are points of tangency with sides AB, BC, CD, DA. After that, we’ll just have to figure out which angles should be same & notice similarity of triangles P’E’F’ and P’G’H’ to prove that.

he showed that $\angle HPG = \angle HPD + \angle GPD = \angle EPB + \angle FPB = \angle EPF$ so equality comes form lemma: if $\frac{\sin{x}}{\sin{(a-x)}} = \frac{\sin{y}}{\sin{(a-y)}}$, then $x = y$ Kayak wrote: Bump ! Can someone answer this ? MathPanda1 wrote: Post #2 wrote: Thus $\frac{\sin \angle HPD}{\sin \angle GPD} = \frac{\sin \angle EPB}{\sin \angle FPB}$, and so we have that $\angle HPD = \angle EPB$ and $\angle GPD = \angle FPB$. Why is this true? Thanks in advance!

Do you mean $\frac{\sin{x}}{\sin{(a-x)}} = \frac{\sin{y}}{\sin{(a-y)}}$?