Given a rectangle $ABCD$, side $AB$ is longer than side $BC$. Find all the points $P$ of the side line $AB$ from which the sides $AD$ and $DC$ are seen from the point $P$ at an equal angle (i.e. $\angle APD = \angle DPC$)

The points $A$ and $P$ are marked on the plane. Consider all such points $B, C $ of this plane that $\angle ABP = \angle MAB$ and $\angle ACP = \angle MAC $, where $M$ is the midpoint of the segment $BC$. Prove that all the circumscribed circles around the triangle $ABC$ for different points $B$ and $C$ pass through some fixed point other than the point $A$. (Alexei Klurman)

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)

Let $ABCD$ be a cyclic quadrilateral such that $\angle ACB = 2\angle CAD$ and $\angle ACD = 2\angle BAC$. Prove that $|CA| = |CB| + |CD|$.

Prove that there is only one triangle whose sides are consecutive natural numbers and one of the angles is twice the other angle.

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 $BD + BA = AC$.

$ABCD$ 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 $\angle DAF = \angle BAF$.

Around the acute-angled triangle $ABC$ ($AC>CB$) a circle is circumscribed, and the point $N$ is midpoint of the arc $ACB$ of this circle. Let the points $A_1$ and $B_1$ be the feet of perpendiculars on the straight line $NC$, drawn from points $A$ and $B$ respectively (segment $NC$ lies inside the segment $A_1B_1$). Altitude $A_1A_2$ of triangle $A_1AC$ and altitude $B_1B_2$ of triangle $B_1BC$ intersect at a point $K$ . Prove that $\angle A_1KN=\angle B_1KM$, where $M$ is midpoint of the segment $A_2B_2$ .

Given a square $ABCD$. Find the locus of points $M$ such that $\angle AMB = \angle CMD$.

Given a triangle $ABC$, show how to construct the point $P$ such that $\angle PAB= \angle PBC= \angle PCA$. Express this angle in terms of $\angle A,\angle B,\angle C$ using trigonometric functions.

Given a triangle $ABC$, let $D$ be a point different from $A$ on the external bisectrix $\ell$ of the angle $BAC$, and let $E$ be an interior point of the segment $AD$. Reflect $\ell$ in the internal bisectrices of the angles $BDC$ and $BEC$ to obtain two lines that meet at some point $F$. Show that the angles $ABD$ and $EBF$ are congruent.

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 point $E$ is located inside a parallelogram $ABCD$ such that $\angle BAE = \angle BCE$. The centers of the circumcircles of the triangles $ABE,ECB, CDE$ and $DAE$ are concyclic.

In the convex quadrilateral $ABCD$ , $\angle ADC = \angle BCD > 90^o$ . Let $E$ be the intersection of the line $AC$ with the line parallel to $AD$ that passes through $B$. Let $F$ be the intersection of line $BD$ with the line parallel to $BC$ passing through $A$. Prove that $EF$ is parallel to $CD$.

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 circles $ S_1 $ and $ S_2 $ intersect at points $ A $ and $ B $. Circle $ S_3 $ externally touches $ S_1 $ and $ S_2 $ at points $ C $ and $ D $ respectively. Let $ PQ $ be a chord cut by the line $ AB $ on circle $ S_3 $, and $ K $ be the midpoint of $ CD $. Prove that $ \angle PKC = \angle QKC $.

A convex quadrilateral $ABCD$ Is placed on the Cartesian plane. Its vertices $A$ and $D$ belong to the negative branch of the graph of the hyperbola $y= 1/x$, the vertices $B$ and $C$ belong to the positive branch of the graph and point $B$ lies at the left of $C$, the segment $AC$ passes through the origin $(0,0)$. Prove that $\angle BAD = \angle BCD$. (I, Voronovich)

Let $ABC$ be a triangle with $\angle A = 90^o, \angle B = 75^o$ and $AB = 2$. The points $P$ and $Q$ on the sides $AC$ and $BC$ respectively are such that $\angle APB = \angle CPQ$ and $\angle BQA = \angle CQP$ . Calculate the measurement of the segment $QA $.

$ABCD$ is a parallelogram. $P$ is a point outside the parallelogram such that angles $\angle PAB$ and $\angle PCB$ have the same value but opposite orientation. Show that $\angle APB = \angle DPC$.

The midpoint of the hypotenuse $AB$ of the right triangle $ABC$ is $K$. The point $M$ on the side $BC$ is taken such that $BM = 2 \cdot MC$. Prove that $\angle BAM = \angle CKM$.

Let the interior point $P$ of the convex quadrilateral $ABCD$ be such that $$|\angle PAD| = |\angle ADP| = |\angle CBP| = |\angle PCB| = |\angle CPD|.$$ Let $O$ be the center of the circumcircle of the triangle $CPD$. Prove that $|OA| = |OB|$.

Let $ABC$ be a triangle with $\angle BAC = 60^o$. Let $E$ be the point on the side $BC$ , such that $2 \angle BAE = \angle ACB$ . Let $D$ be the second intersection of $AB$ and the circumcircle of the triangle $AEC$ and $P$ be the second intersection of $CD$ and the circumcircle of the triangle $DBE$. Calculate the angle $\angle BAP$.

In a triangle $ABC$, let $AA', BB'$ and $CC'$ be the bisectors. Suppose $A'B' \cap CC' =P$ and $A'C' \cap BB'= Q$. Prove that $\angle PAC = \angle QAB$.

Given a triangle $ABC$, the perpendicular bisector of the side $AC$ intersects the angle bisector of the triangle $AK$ at the point $P$, $M$ - such a point that $\angle MAC = \angle PCB$, $\angle MPA = \angle CPK$, and points $M$ and $K$ lie on opposite sides of the line $AC$. Prove that the line $AK$ bisects the segment $BM$. (Anton Trygub)

We consider the quadrilateral $ABCD$, with $AD = CD = CB$ and $AB \parallel CD$. Points $E$ and $F$ belong to the segments $CD$ and $CB$ so that angles $\angle ADE = \angle AEF$. Prove that: a) $4CF \le CB$ , b) if $4CF = CB$, then $AE$ is the bisector of the angle $\angle DAF$.