Found problems: 361
On the sides $AB$, $BC$ and $CA$ of the isosceles triangle $ABC$ with the vertex at the point $B$ marked the points $M$, $D$ and $K$ respectively so that $AM = 2DC$ and $\angle AMD = \angle KDC$. Prove that $MD = KD$.
Given a convex quadrilateral $ KLMN $, in which $ \angle NKL = {{90} ^ {\circ}} $. Let $ P $ be the midpoint of the segment $ LM $. It turns out that $ \angle KNL = \angle MKP $. Prove that $ \angle KNM = \angle LKP $.
Let $ABC$ be an acute triangle, with $AB <AC$. Let $M$ be the midpoint of $AB$, $H$ the foot of the altitude from $A$, and $Q$ be point on side $AC$ such that $\angle ABQ = \angle BCA$. Show that the circumcircles of the triangles $ABQ$ and $BHM$ are tangent.
$ABCD$ is a parallelogram. A point $M$ is found on the side $AB$ or its extension such that $\angle MAD = \angle AMO$ where $O$ is the intersection point of the diagonals of the parallelogram. Prove that $MD = MG$.
(M Smurov)
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 $.
The diagonals $AC$ and $BD$ of a convex quadrilateral $ABCD$ intersect in point $M$. The angle bisector of $\angle ACD$ intersects the ray $\overrightarrow{BA}$ in point $K$. If
$MA.MC+MA.CD=MB.MD$,
prove that $\angle BKC=\angle CDB$.
In a triangle $ABC$, let $K$ be a point on the median $BM$ such that $CM = CK$. It turned out that $\angle CBM = 2\angle ABM$. Show that $BC = KM$.
An acute-angled triangle $ABC$ has two altitudes $BE$ and $CF$. The circle with diameter $AC$ intersects the segment $BE$ at point $P$. A circle with diameter $AB$ intersects the segment $CF$ at point $Q$ and the extension of this altitude at point $Q'$. Prove that $\angle PQ'Q = \angle PQB$.
A scalene acute triangle $ABC$ is drawn on the plane, in which $BC$ is the longest side. Points $P$ and $D$ are constructed, the first inside $ABC$ and the second outside, so that $\angle ABC = \angle CBD$, $\angle ACP = \angle BCD$ and that the area of triangle $ABC$ is equal to the area of quadrilateral $BPCD$. Prove that triangles $BCD$ and $ACP$ are similar.
The circles $\Gamma_1$ and $\Gamma_2$ intersect at $D$ and $P$. The common tangent line of the two circles closest to point $D$ touches $\Gamma_1$ in A and $\Gamma_2$ in $B$. The line $AD$ intersects $\Gamma_2$ for the second time in $C$. Let $M$ be the midpoint of line segment $BC$. Prove that $\angle DPM = \angle BDC$.
Each pair of opposite sides of a convex hexagon has the following property: the distance between their midpoints is equal to $\dfrac{\sqrt{3}}{2}$ times the sum of their lengths. Prove that all the angles of the hexagon are equal.
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 $\angle CPM = \angle DPL$.
Let $ABCD$ be an inscribed quadrilateral. Denote the midpoints of the sides $AB, BC, CD$ and $DA$ through $M, L, N$ and $K$, respectively. It turned out that $\angle BM N = \angle MNC$. Prove that:
i) $\angle DKL = \angle CLK$.
ii) in the quadrilateral $ABCD$ there is a pair of parallel sides.
Let $BM$ be the median of the triangle $ABC$, in which $AB> BC$. Point $P$ is chosen so that $AB \parallel PC$ and $PM \perp BM$. Prove that $\angle ABM = \angle MBP$.
(Mikhail Standenko)
In triangle $ABC$, point $M$ is the midpoint of $BC, P$ is the intersection point of the tangents at points $B$ and $C$ of the circumscribed circle, $N$ is the midpoint of the segment $MP$. The segment $AN$ intersects the circumscribed circle at point $Q$. Prove that $\angle PMQ = \angle MAQ$.
Consider a convex quadrilateral $ABCD$ with $AB=CB$ and $\angle ABC +2 \angle CDA = \pi$ and let $E$ be the midpoint of $AC$. Show that $\angle CDE =\angle BDA$.
Paolo Leonetti
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 $OP = OQ$.
Let $M$ be the point of intersection of the diagonals $AC$ and $BD$ of the convex quadrilateral $ABCD$. Let$ K$ be the intersection point of the extension of side $AB$ (from $A$) with the bisector of the $\angle ACD$. If $MA \cdot MC + MA \cdot CD = MB\cdot MD$ , prove that $\angle BKC = \angle CDB$.
Two circles intersect at points $P$ and $Q$. The straight line intersects these circles at points $A$, $B$, $C$, $D$, as shown in fig. . Prove that $\angle APB = \angle CQD$.
[img]https://cdn.artofproblemsolving.com/attachments/1/a/a581e11be68bbb628db5b5b8e75c7ff6e196c5.png[/img]
Let $ABCD$ be a quadrilateral with $\angle A = \angle B = 90^o$, $AB = AD$. Denote $E$ as the midpoint of $AD$, suppose that $CD = BC + AD$, $AD > BC$. Prove that $\angle ADC = 2\angle ABE$.
In a triangle $ABC$ the point $D$ is the intersection of the interior angle bisector of $\angle BAC$ and side $BC$. Let $P$ be the second intersection point of the exterior angle bisector of $\angle BAC$ with the circumcircle of $\angle ABC$. A circle through $A$ and $P$ intersects line segment $BP$ internally in $E$ and line segment $CP$ internally in $F$. Prove that $\angle DEP = \angle DFP$.
In the triangle $ABC$, $AD$ is altitude, $M$ is the midpoint of $BC$. It is known that $\angle BAD = \angle DAM = \angle MAC$. Find the values of the angles of the triangle $ABC$
Let $ABC$ be a triangle with right $\angle A$. Consider points $D \in (AC)$ and $E \in (BD)$ such that $\angle ABC = \angle ECD = \angle CED$. Prove that $BE = 2 \cdot AD$
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)
On the sides $AB, BC$ and $CA$ of a triangle $ABC$ are located the points $P, Q$ and $R$ respectively, such that $BQ = 2QC, CR = 2RA$ and $\angle PRQ = 90^o$. Show that $\angle APR =\angle RPQ$.