In $\triangle ABC$, points $P,Q$ satisfy $\angle PBC = \angle QBA$ and $\angle PCB = \angle QCA$, $D$ is a point on $BC$ such that $\angle PDB=\angle QDC$. Let $X,Y$ be the reflections of $A$ with respect to lines $BP$ and $CP$, respectively. Prove that $DX=DY$. [img]https://cdn.artofproblemsolving.com/attachments/a/7/f208f1651afc0fef9eef4c68ba36bf77556058.jpg[/img]

Points $K$ and $L$ are taken on the sides $BC$ and $CD$ of a square $ABCD$ so that $\angle AKB = \angle AKL$. Find $\angle KAL$.

In the quadrilateral $ABCD$, angles $B$ and $C$ are equal to $120^o$, $AB = CD = 1$, $CB = 4$. Find the length $AD$.

The vertex $F$ of the parallelogram $ACEF$ lies on the side $BC$ of parallelogram $ABCD$. It is known that $AC = AD$ and $AE = 2CD$. Prove that $\angle CDE = \angle BEF$.

Let $\Omega$ be the circumcircle of an acute-angled triangle $ABC$. A point $D$ is chosen on the internal bisector of $\angle ACB$ so that the points $D$ and $C$ are separated by $AB$. A circle $\omega$ centered at $D$ is tangent to the segment $AB$ at $E$. The tangents to $\omega$ through $C$ meet the segment $AB$ at $K$ and $L$, where $K$ lies on the segment $AL$. A circle $\Omega_1$ is tangent to the segments $AL, CL$, and also to $\Omega$ at point $M$. Similarly, a circle $\Omega_2$ is tangent to the segments $BK, CK$, and also to $\Omega$ at point $N$. The lines $LM$ and $KN$ meet at $P$. Prove that $\angle KCE = \angle LCP$. Poland

Consider in a square $[ABCD]$ a point $E$ on the side $AB$, different from $A$ and $B$. On the side $BC$ consider the point $F$ such that $\angle AED = \angle DEF$ . Prove that $EF = AE + FC$.

A point $P$ was chosen on the smaller arc $BC$ of the circumcircle of the acute-angled triangle $ABC$. Points $R$ and $S$ on the sides$ AB$ and $AC$ are respectively selected so that $CPRS$ is a parallelogram. Point $T$ on the arc $AC$ of the circumscribed circle of $\vartriangle ABC$ such that $BT \parallel CP$. Prove that $\angle TSC = \angle BAC$. (Anton Trygub)

Given a rectangular parallelepiped $ABCDA_1B_1C_1D_1$. Let the points $E$ and $F$ be the feet of the perpendiculars drawn from point $A$ on the lines $A_1D$ and $A_1C$, respectively, and the points $P$ and $Q$ be the feet of the perpendiculars drawn from point $B_1$ on the lines $A_1C_1$ and $A_1C$, respectively. Prove that $\angle EFA = \angle PQB_1$

Given is a parallelogram $ABCD$ with $\angle A < 90^o$ and $|AB| < |BC|$. The angular bisector of angle $A$ intersects side $BC$ in $M$ and intersects the extension of $DC$ in $N$. Point $O$ is the centre of the circle through $M, C$, and $N$. Prove that $\angle OBC = \angle ODC$. [asy] unitsize (1.2 cm); pair A, B, C, D, M, N, O; A = (0,0); B = (2,0); D = (1,3); C = B + D - A; M = extension(A, incenter(A,B,D), B, C); N = extension(A, incenter(A,B,D), D, C); O = circumcenter(C,M,N); draw(D--A--B--C); draw(interp(D,N,-0.1)--interp(D,N,1.1)); draw(A--interp(A,N,1.1)); draw(circumcircle(M,C,N)); label("$\circ$", A + (0.45,0.15)); label("$\circ$", A + (0.25,0.35)); dot("$A$", A, SW); dot("$B$", B, SE); dot("$C$", C, dir(90)); dot("$D$", D, dir(90)); dot("$M$", M, SE); dot("$N$", N, dir(90)); dot("$O$", O, SE); [/asy]

In triangle $ABC$, angle $C$ is a right angle. Found on the side $AC$ point $D$, and on the segment $BD$, point $K$ such that $\angle ABC = \angle KAD =\angle AKD$. Prove that $BK = 2DC$.

A point $M$ is chosen inside the square $ABCD$ in such a way that $\angle MAC = \angle MCD = x$ . Find $\angle ABM$.

The diagonals of quadrilateral $ABCD$ intersect in point $E$. Given that $|AB| =|CE|$, $|BE| = |AD|$, and $\angle AED = \angle BAD$, determine the ratio $|BC|:|AD|$.

Let $O$ be the center of the circle of the triangle $ABC$ with $AB \ne AC$. Furthermore, let $S$ and $T$ be points on the rays $AB$ and $AC$, such that $\angle ASO = \angle ACO$ and $\angle ATO = \angle ABO$. Show that $ST$ bisects the segment $BC$.

Let $E$ be a point lies inside the parallelogram $ABCD$ such that $\angle BCE = \angle BAE$. Prove that the circumcenters of triangles $ABE,BCE,CDE,DAE$ are concyclic.

Given a triangle $ABC$, in which the angle $B$ is three times the angle $C$. On the side $AC$, point $D$ is chosen such that the angle $BDC$ is twice the angle $C$. Prove that $BD + BA = AC$.

Let $AD$ be the bisector of angle $A$ of the triangle $ABC$. One considers the points M, N on the half-lines $(AB$ and $(AC$, respectively, such that $\angle MDA = \angle B$ and $\angle NDA = \angle C$. Let $AD \cap MN=\{P\}$. Prove that: $$AD^3 = AB \cdot AC\cdot AP$$

In parallelogram $ABCD$ with acute angle $A$ a point $N$ is chosen on the segment $AD$, and a point $M$ on the segment $CN$ so that $AB = BM = CM$. Point $K$ is the reflection of $N$ in line $MD$. The line $MK$ meets the segment $AD$ at point $L$. Let $P$ be the common point of the circumcircles of $AMD$ and $CNK$ such that $A$ and $P$ share the same side of the line $MK$. Prove that $\angle CPM = \angle DPL$.

The lateral sidelines $AB$ and $CD$ of trapezoid $ABCD$ meet at point $S$. The bisector of angle $ASC$ meets the bases of the trapezoid at points $K$ and $L$ ($K$ lies inside segment $SL$). Point $X$ is chosen on segment $SK$, and point $Y$ is selected on the extension of $SL$ beyond $L$ such a way that $\angle AXC - \angle AYC = \angle ASC$. Prove that $\angle BXD - \angle BYD = \angle BSD$.

Let $ABCD$ be quadrilateral inscribed in a circle. Let $M$ be the midpoint of the segment $BD$. If the tangents of the circle at $ B$, and at $D$ are also concurrent with the extension of $AC$, prove that $\angle AMD = \angle CMD$.

Define acute triangle $ABC$ with circumcircle $\omega.$ Let $Q$ be the midpoint of minor arc $BC$ in $\omega$ and let $Q'$ be the reflection of $Q$ over $BC.$ If the circle with diameter $BC$ is tangent to the external angle bisector of $\angle BAC$ at $P,$ show $\angle BPQ' = \angle CPA.$ [i]Proposed by Evan Chang (squareman), USA[/i] [img]https://cdn.artofproblemsolving.com/attachments/8/1/6333de3458f913477c75882896a40a48cd7ef7.png[/img]

Let $O$ be the circumcenter of the triangle $ABC, A'$ be a point symmetric of $A$ wrt line $BC, X$ is an arbitrary point on the ray $AA'$ ($X \ne A$). Angle bisector of angle $BAC$ intersects the circumcircle of triangle $ABC$ at point $D$ ($D \ne A$). Let $M$ be the midpoint of the segment $DX$. A line passing through point $O$ parallel to $AD$, intersects $DX$ at point $N$. Prove that angles $BAM$ and $CAN$ angles are equal.

$ABCD$ is a convex quadrilateral, and the point $K$ is a point on side $AB$, where $\angle KDA=\angle BCD$, let $L$ be a point on the diagonal $AC$, where $KL$ is parallel to $BC$. Prove that $$\angle KDB=\angle LDC.$$ [i](tatari/nightmare)[/i]

Given a quadrangle $ABCD$ and a point $M$ inside it such that $ABMD$ is a parallelogram. $ \angle CBM = \angle CDM$. Prove that the $ \angle ACD = \angle BCM$.

Two trapezoid angles and diagonals are respectively equal. Is it true that such are the trapezoid equal?

Let $ABC$ be a triangle with $AH$ altitude. The point $K$ is chosen on the segment $AH$ as follows such that $AH =3KH$. Let $O$ be the center of the circle circumscribed around by triangle $ABC, M$ and $N$ be the midpoints of $AC$ and AB respectively. Lines $KO$ and $MN$ intersect at the point $Z$, a perpendicular to $OK$ passing through point $Z$ intersects lines $AC$ and $AB$ at points $X$ and $Y$ respectively. Prove that $\angle XKY =\angle CKB$.