A trapezoid $ABCD$ with perpendicular diagonals $AC$ and $BD$ is inscribed in a circle $k$. Let $k_a$ and $k_c$ respectively be the circles with diameters $AB$ and $CD$. Compute the area of the region which is inside the circle $k$, but outside the circles $k_a$ and $k_c$.

In the acute-angled triangle $ABC$, the segment $AP$ was drawn and the center was marked $O$ of the circumscribed circle. The circumcircle of triangle $ABP$ intersects the line $AC$ for the second time at point $X$, the circumcircle of the triangle $ACP$ intersects the line $AB$ for the second time at the point $Y$. Prove that the lines $XY$ and $PO$ are perpendicular if and only if $P$ is the foor of the bisector of the triangle $ABC$.

Circles $C$ and $C'$ intersect at $O$ and $X$. A circle center $O$ meets $C$ at $Q$ and $R$ and meets $C'$ at $P$ and $S$. $PR$ and $QS$ meet at $Y$ distinct from $X$. Show that $\angle YXO = 90^o$.

Let $H$ be the orthocenter of the acute triangle $ABC$. Let $H_a$ be the foot of the perpendicular from $A$ to $BC$ and let the line through $H$ parallel to $BC$ intersect the circle with diameter $AH_a$ in the points $P_a$ and $Q_a$. Similarly, we define the points $P_b, Q_b$ and $P_c,Q_c$. Show that the six points $P_a,Q_a,P_b,Q_b,P_c,Q_c$ lie on a common circle.

Let $ABC$ be a triangle inscribed in a circle of center $O$ and $P$ be a point on the arc $AB$, that does not contain $C$. The perpendicular drawn fom $P$ on line $BO$ intersects $AB$ at $S$ and $BC$ at $T$. The perpendicular drawn from $P$ on line $AO$ intersects $AB$ at $Q$ and $AC$ at $R$. Prove that: a) $PQS$ is an isosceles triangle b) $PQ^2=QR= ST$

The orthocenter of an acute triangle $ABC$ is $H$. Let $K$ and $P$ be the midpoints of lines $BC$ and $AH$, respectively. The angle bisector drawn from the vertex $A$ of the triangle $ABC$ intersects with line $KP$ at $D$. Prove that $HD\perp AD$.

Given an acute triangle $ABC, H$ is the foot of the altitude drawn from the point $A$ on the line $BC, P$ and $K \ne H$ are arbitrary points on the segments $AH$ and$ BC$ respectively. Segments $AC$ and $BP$ intersect at point $B_1$, lines $AB$ and $CP$ at point $C_1$. Let $X$ and $Y$ be the projections of point $H$ on the lines $KB_1$ and $KC_1$, respectively. Prove that points $A, P, X$ and $Y$ lie on one circle.

Given a circle $\gamma$ and a point $P$ inside it, find the maximum and minimum value of the sum of the lengths of two perpendicular chords of $\gamma$ passing through $P$.

Let $ABC$ be a triangle. $(K)$ is an arbitrary circle tangent to the lines $AC,AB$ at $E, F$ respectively. $(K)$ cuts $BC$ at $M,N$ such that $N$ lies between $B$ and $M$. $FM$ intersects $EN$ at $I$. The circumcircles of triangles $IFN$ and $IEM$ meet each other at $J$ distinct from $I$. Prove that $IJ$ passes through $A$ and $KJ$ is perpendicular to $IJ$. Trần Quang Hùng

On the legs $AC, BC$ of a right triangle $\vartriangle ABC$ select points $M$ and $N$, respectively, so that $\angle MBC = \angle NAC$. The perpendiculars from points $M$ and $C$ on the line $AN$ intersect $AB$ at points $K$ and $L$, respectively. Prove that $KL=LB$. (O. Clurman)

Let $ABCD$ be a convex quadrilateral with perpendicular diagonals. If $AB = 20, BC = 70$ and $CD = 90$, then what is the value of $DA$?

Consider an acute triangle $ABC$ and it's circumcircle $\omega$. With center $A$, we construct a circle $\gamma$ that intersects arc $AB$ of circle $\omega$ , that doesn't contain $C$, at point $D$ and arc $AC$ , that doesn't contain $B$, at point $E$. Suppose that the intersection point $K$ of lines $BE$ and $CD$ lies on circle $\gamma$. Prove that line $AK$ is perpendicular on line $BC$.

As shown in the figure, it is known that the quadrilateral $ABCD$ satisfies $\angle ADB = \angle ACB = 90^o$. Suppose $AC$ and $BD$ intersect at point $P$, point $R$ lies on $CD$ and $RP \perp AB$. $M$ and $N$ are the midpoints of $AB$ and $CD$ respectively. Point $K$ is a point on the extension line of $NM$, the circumscribed circles of $\vartriangle DKC$ and $\vartriangle AKB$ intersect at point $S$. Prove that $KS \perp SR$. [img]https://cdn.artofproblemsolving.com/attachments/5/d/fc0a391f8ebcdee792e9b226cbf55a058251a1.png[/img]

Prove that the area of a unit cube's projection on any plane equals the length of its projection on the perpendicular to this plane.

Suppose a triangle $\vartriangle ABC$ with $\angle C = 90^o$ is given. Let $D$ be the midpoint of $AC$, and let $E$ be the foot of the altitude through $C$ on $BD$. Show that the tangent in $C$ of the circumcircle of $\vartriangle AEC$ is perpendicular to $AB$.

From all three vertices of triangle $ABC$, we set perpendiculars to the exterior and interior of the other vertices angle bisectors. Prove that the sum of the squares of the segments thus obtained is exactly $2 (a^2 + b^2 + c^2)$, where $a, b$, and $c$ denote the lengths of the sides of the triangle.

Let $ABC$ be a triangle with $AC \ne BC$. In triangle $ABC$, let $G$ be the centroid, $I$ the incenter and O Its circumcenter. Prove that $IG$ is parallel to $AB$ if, and only if, $CI$ is perpendicular on $IO$.

Given an isosceles triangle $ABC$ with $AB = AC$. Let $O$ be the circumcenter of $ABC$, $D$ the midpoint of $AB$ and $E$ the centroid of $ACD$. Prove that $CD \perp EO$.

On side $AB$ of triangle $ABC$, point $D$ is selected. Circle circumscribed around triangle $BCD$, intersects side $AC$ at point $M$, and the circumcircle of triangle $ACD$ intersects the side $BC$ at point $ N$ ($M,N \ne C$). Let $O$ be the circumcenter of the triangle $CMN$. Prove that line $OD$ is perpendicular to side $AB$.

$ABCD$ is a parallelogram. Take $E$ on the line $AB$ so that $BE = BC$ and $B$ lies between $A$ and $E$. Let the line through $C$ perpendicular to $BD$ and the line through $E$ perpendicular to $AB$ meet at $F$. Show that $\angle DAF = \angle BAF$.

In an acute triangle $ABC$ let $K,L,$ and $M$ be the midpoints of sides $AB,BC,$ and $CA,$ respectively. From each of $K,L,$ and $M$ drop two perpendiculars to the other two sides of the triangle; e.g., drop perpendiculars from $K$ to sides $BC$ and $CA,$ etc. The resulting $6$ perpendiculars intersect at points $Q,S,$ and $T$ as in the figure to form a hexagon $KQLSMT$ inside triangle $ABC.$ Prove that the area of this hexagon $KQLSMT$ is half of the area of the original triangle $ABC.$ [asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra; diagram by adihaya*/ import graph; size(10cm); 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 = 11.888712276357234, xmax = 17.841346447833423, ymin = 10.61620970860601, ymax = 15.470685507068502; /* image dimensions */ pen zzttqq = rgb(0.6,0.2,0.); pen qqwuqq = rgb(0.,0.39215686274509803,0.); pair A = (12.488234161849352,12.833838721895551), B = (16.50823416184936,15.093838721895553), C = (16.28823416184936,11.353838721895551), K = (14.498234161849355,13.963838721895552), L = (16.39823416184936,13.223838721895552), M = (14.388234161849356,12.093838721895551), D = (13.615830174638527,13.467760858438725), F = (15.75135711740064,11.562938202365055), G = (15.625830174638523,14.597760858438724), H = (16.435061748056253,13.849907687412797), T = (14.02296781802369,12.74356027153236), Q = (16.032967818023693,13.873560271532357), O = (16.325061748056253,11.979907687412794); draw(A--B--C--cycle, zzttqq); draw((13.426050287639166,13.361068683160477)--(13.532742462917415,13.171288796161116)--(13.722522349916774,13.277980971439364)--D--cycle, qqwuqq); draw((14.054227993863618,12.223925334689998)--(14.133240861538676,12.426796211152979)--(13.930369985075695,12.505809078828037)--(13.851357117400637,12.302938202365056)--cycle, qqwuqq); draw((16.337846386707046,12.19724654447628)--(16.12050752964356,12.210031183127075)--(16.107722890992765,11.992692326063588)--O--cycle, qqwuqq); draw((15.830369985075697,11.765809078828037)--(15.627499108612716,11.844821946503092)--(15.54848624093766,11.641951070040111)--F--cycle, qqwuqq); draw((15.436050287639164,14.491068683160476)--(15.542742462917412,14.301288796161115)--(15.73252234991677,14.407980971439365)--G--cycle, qqwuqq); draw((16.217722890992764,13.86269232606359)--(16.20493825234197,13.645353469000101)--(16.42227710940546,13.63256883034931)--H--cycle, qqwuqq); Label laxis; laxis.p = fontsize(10); xaxis(xmin, xmax, Ticks(laxis, Step = 1., Size = 2, NoZero),EndArrow(6), above = true); yaxis(ymin, ymax, Ticks(laxis, Step = 1., Size = 2, NoZero),EndArrow(6), above = true); /* draws axes; NoZero hides '0' label */ /* draw figures */ draw(A--B, zzttqq); draw(B--C, zzttqq); draw(C--A, zzttqq); draw(M--D); draw(K--(13.851357117400637,12.302938202365056)); draw(F--L); draw(L--G); draw(K--H); draw(M--O); /* dots and labels */ dot(A,dotstyle); label("$A$", (12.52502834296331,12.93568440300881), NE * labelscalefactor); dot(B,dotstyle); label("$B$", (16.548187989892043,15.193580123223922), NE * labelscalefactor); dot(C,dotstyle); label("$C$", (16.332661580235147,11.457789022504372), NE * labelscalefactor); dot(K,linewidth(3.pt) + dotstyle); label("$K$", (14.536608166427676,14.02357961365791), NE * labelscalefactor); dot(L,linewidth(3.pt) + dotstyle); label("$L$", (16.43529320388129,13.28463192340569), NE * labelscalefactor); dot(M,linewidth(3.pt) + dotstyle); label("$M$", (14.433976542781535,12.155684063298134), NE * labelscalefactor); dot(D,linewidth(3.pt) + dotstyle); dot((13.851357117400637,12.302938202365056),linewidth(3.pt) + dotstyle); dot(F,linewidth(3.pt) + dotstyle); dot(G,linewidth(3.pt) + dotstyle); dot(H,linewidth(3.pt) + dotstyle); dot((15.922967818023695,12.003560271532354),linewidth(3.pt) + dotstyle); label("$S$", (15.96318773510904,12.063315602016607), NE * labelscalefactor); dot(T,linewidth(3.pt) + dotstyle); label("$T$", (14.064502697655428,12.802263292268826), NE * labelscalefactor); dot(Q,linewidth(3.pt) + dotstyle); label("$Q$", (16.076082521119794,13.931211152376383), NE * labelscalefactor); dot(O,linewidth(3.pt) + dotstyle); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy]

Let $ABC$ be a triangle with $AB \ne AC$. The incirle of triangle $ABC$ is tangent to $BC, CA, AB$ at $D, E, F$, respectively. The perpendicular line from $D$ to $EF$ intersects $AB$ at $X$. The second intersection point of circumcircles of triangles $AEF$ and $ABC$ is $T$. Prove that $TX \perp T F$

Quadrangle $ABCD$ is inscribed in a circle. The perpendicular from the vertex $C$ on the bisector of $\angle ABD$ intersects the line $AB$ at the point $C_1$. The perpendicular from the vertex $B$ on the bisector of $\angle ACD$ intersects the line $CD$ at the point $B_1$. Prove that $B_1C_1 \parallel AD$.

Given four points $A,$ $B,$ $C,$ $D$ on a circle such that $AB$ is a diameter and $CD$ is not a diameter. Show that the line joining the point of intersection of the tangents to the circle at the points $C$ and $D$ with the point of intersection of the lines $AC$ and $BD$ is perpendicular to the line $AB.$

As shown in the figure, it is known that $BC= AC$ in $\vartriangle ABC$, $M$ is the midpoint of $AB$, points $D$, $E$ lie on $AB$ such that $\angle DCE= \angle MCB$, the circumscribed circle of $\vartriangle BDC$ and the circumscribed circle of $\vartriangle AEC$ intersect at point $F $(different from point $C$), point $H$ lies on $AB$ such that the straight line $CM$ bisects the line segment $HF$. Let the circumcenters of $\vartriangle HFE$ and $\vartriangle BFM$ be $O_1$, $O_2$ respectively. Prove that $O_1O_2 \perp CF$. [img]https://cdn.artofproblemsolving.com/attachments/8/c/62d4ecbc18458fb4f2bf88258d5024cddbc3b0.jpg[/img]