Proof of part (a) Lemma If $ABCD$ is a cyclic quadrilateral with perpendicular diagonals at $P$ then if $M$ is the midpoint of $AB$ then let $PM$ meet $CD$ in $M'$ the $M'$ is the projection from $P$ to $CD$

PROOF

Coming back to the problem

i) By Brahmagupta's Theorem, $P, E, Y$, $P, F, Z$, $P, G, W$ and $P, H, X$ are colinear. Since $EH\parallel BD\parallel FG$, $EF\parallel AC\parallel GH$ and $AC\perp BD$, we see that $EFGH$ is rectangle. Now let $M=EF\cap GH$ therefore $ME=MF=MG=MH$. Since $\angle EWG=\angle EYG=\angle FXH=\angle FZH=90^{\circ}$, we see that $MW=MX=MY=MZ=ME$ hence $E, F, G, H, W, X, Y, Z$ are concyclic. ii) It suffices to find $\frac{EG}{2}=\frac{1}{2}\sqrt{EF^2+EG^2}=\frac{1}{4}\sqrt{AC^2+BD^2}$. Let $O$ be the center of $\odot(ABCD)$ and let $K, L$ be the midpoints of $AC, BD$. Then we find \begin{align*}AC^2+BD^2&= 4(AK^2+BL^2)\\ &=4((R^2-OK^2)+(R^2-OL^2))\\ &=8R^2-4OP^2\end{align*}Hence the final answer is $\dfrac{\sqrt{2R^2-OP^2}}{2}$.

Part 1) By Brahmagupta's Theorem, $P, E, Y$, $P, F, Z$, $P, G, W$ and $P, H, X$ are colinear. Be simple length chasing, we get $PE\times PY = PF\times PZ= PG\times PW = PH \times PX= 1/2 \times PA \times PC$

anantmudgal09 wrote: Let $ABCD$ be a cyclic quadrilateral inscribed in circle $\Omega$ with $AC \perp BD$. Let $P=AC \cap BD$ and $W,X,Y,Z$ be the projections of $P$ on the lines $AB, BC, CD, DA$ respectively. Let $E,F,G,H$ be the mid-points of sides $AB, BC, CD, DA$ respectively. (a) Prove that $E,F,G,H,W,X,Y,Z$ are concyclic. (b) If $R$ is the radius of $\Omega$ and $d$ is the distance between its centre and $P$, then find the radius of the circle in (a) in terms of $R$ and $d$. Clearly $E,P,Y$ are collinear(By Brahmagupta's Theorem).. Also By some angle chasing we get $EFGH$ as a rectangle.Clearly $\angle EWG =90$ thus $W$ lies on the circumcircle of the rectangle.Thus $E,F,G,H,W,X,Y,Z$ are concyclic. Note Radius of this circle = Diagonal of the rectangle$=\frac{1}{4}\sqrt{AC^2+BD^2}$ We drop perpendiculars from $O$ to $AC$ and $BD$ at $S$ and $T$ .. By Pythagoras and noting $OSPT$ is a rectangle, We get $Radius=\dfrac{\sqrt{2R^2-d^2}}{2}$

Just a different solution for $a)$ Let $BC\cap AD=N$ and $AB\cap CD=M$. Let $O$ be the center of $\odot(ABCD)$. Then $\{P,O\}$ Are pairs of Isogonal Conjugates WRT $\triangle CND$ and $\triangle AMD$. So, the Pedal Circles of $P$ and $O$ WRT $\triangle CND$ is same. So, $\{X,F,Y,G,H,Z\}$ lies on a circle. So as the Pedal Circles of $\{P,O\}$ WRT $\triangle AMD$ will be same $\implies \{W,E,Z,H,Y,G\}$ lies on a circle. Now as three unique points determine a unique circle, hence, $\{E,F,G,H,W,X,Y,Z\}$ are concyclic.

Part (A) can be finished even more easily. Let $O$ be the center of $\Omega$. Notice that $P$ and $O$ are isogonal conjugates with respect to quadrilateral $ABCD$, so it is well-known that their pedal circles coincide.