Let $ABC$ be an acute triangle and $D$ a point on the side $BC$ such that $\angle BAD = \angle DAC$. The circumcircles of the triangles $ABD$ and $ACD$ intersect the segments $AC$ and $AB$ at $E$ and $F$, respectively. The internal bisectors of $\angle BDF$ and $\angle CDE$ intersect the sides $AB$ and $AC$ at $P$ and $Q$, respectively. Points $X$ and $Y$ are chosen on the side $BC$ such that $PX$ is parallel to $AC$ and $QY$ is parallel to $AB$. Finally, let $Z$ be the point of intersection of $BE$ and $CF$. Prove that $ZX = ZY$.

In triangle $ABC$, point $M$ is the midpoint of $BC$, $P$ the point of intersection of the tangents at points $B$ and $C$ of the circumscribed circle of $ABC$, $N$ is the midpoint of the segment $MP$. The segment $AN$ meets the circumcircle $ABC$ at the point $Q$. Prove that $\angle PMQ = \angle MAQ$.

In $\triangle ABC$ we have that $CC_{1}$ is an angle bisector. The points $P\in C_{1}B$, $Q\in BC$, $R\in AC$, $S\in AC_{1}$ satisfy $C_{1}P=PQ=QC$ and $CR=RS=SC_{1}$. Prove that $CC_{1}$ bisects $\angle SCP$.

Let $ABC$ be an acute triangle, and let $D$ be the foot of the altitude from $A$. The circle with centre $A$ passing through $D$ intersects the circumcircle of triangle $ABC$ in $X$ and $Y$ , in such a way that the order of the points on this circumcircle is: $A, X, B, C, Y$ . Show that $\angle BXD = \angle CYD$.

Let $ABC$ be a triangle inscribed in the circle $K$ and consider a point $M$ on the arc $BC$ that do not contain $A$. The tangents from $M$ to the incircle of $ABC$ intersect the circle $K$ at the points $N$ and $P$. Prove that if $\angle BAC = \angle NMP$, then triangles $ABC$ and $MNP$ are congruent. Valentin Vornicu [hide= about Romania JBMO TST 2004 in aops]I found the Romania JBMO TST 2004 links [url=https://artofproblemsolving.com/community/c6h5462p17656]here [/url] but they were inactive. So I am asking for solution for the only geo I couldn't find using search. The problems were found [url=https://artofproblemsolving.com/community/c6h5135p16284]here[/url].[/hide]

The segments $AC$ and $BD$ are perpendicular, and $AC$ is twice as large as $BD$ and intersects $BD$ in it in the midpoint. Find the value of the angle $BAD$, if we know that $\angle CAD = \angle CDB$. (Gregory Filippovsky)

Given the rectangle $ABCD$. The points $E ,F$ lie on the segments $(BC) , (DC)$ respectively, such that $\angle DAF = \angle FAE$. Proce that if $DF + BE = AE$ then $ABCD$ is square.

In a convex quadrilateral $ABCD$ the angles $\angle A$ and $\angle C$ are equal and the bisector of $\angle B$ passes through the midpoint of the side $CD$. If it is known that $CD = 3AD$, calculate $\frac{AB}{BC}$.

Let $P$ be a point inside a triangle $ABC$ such that $\angle PAB = \angle PCA$ and $\angle PAC = \angle PBA$. If $O \ne P$ is the circumcenter of $\triangle ABC$, prove that $\angle APO$ is right.

Show that if equiangular hexagon has sides $a, b, c, d, e, f$ in order then $a - d = e - b = c - f$.

Let $\vartriangle ABC$ be a triangle with circumcenter $O$. Let $ P$ and $Q$ be internal points on the sides $AB$ and $AC$ respectively such that $\angle POB = \angle ABC$ and $\angle QOC = \angle ACB$. Show that the reflection of line $BC$ over line $PQ$ is tangent to the circumcircle of triangle $\vartriangle APQ$.

A (convex) trapezoid $ABCD$ is good, if it is inscribed in a circle, sides $AB$ and $CD$ are the bases and $CD$ is shorter than $AB$. For a good trapezoid $ABCD$ the following terms are defined: $\bullet$ The parallel to $AD$ passing through $B$ intersects the extension of side $CD$ at point $S$. $\bullet$ The two tangents passing through $S$ on the circumircle of the trapezoid touch the circle at $E$ and $F$, where $E$ lies on the same side of the straight line $CD$ as $A$. Give the simplest possible equivalent condition (expressed in side lengths and / or angles of the trapezoid) so that with a good trapezoid $ABCD$ the two angles $\angle BSE$ and $\angle FSC$ have the same measure. (Walther Janous)

$ABCD$ 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]

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]

In the trapezoid $ABCD, AB // CD$ and the diagonals intersect at $O$. The points $P, Q$ are on $AD, BC$ respectively such that $\angle AP B = \angle CP D$ and $\angle AQB = \angle CQD$. Show that $OP = OQ$.

a) Does there exist a quadrilateral with (both of) the following properties: three of its edges are of the same length, but the fourth one is different, and three of its angles are equal, but the fourth one is different? b) Does there exist a pentagon with (both of) the following properties: four of its edges are of the same length, but the fifth one is different, and four of its angles are equal, but the fifth one is different?