Found problems: 361
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 $\angle AMD = \angle CMD$.
In a trapezium with bases $AD$ and $BC$, let $P$ and $Q$ be the middles of diagonals $AC$ and $BD$ respectively. Prove that if $\angle DAQ = \angle CAB$ then $\angle PBA = \angle DBC$.
In a triangle $[ABC]$, $\angle C = 2\angle A$. A point $D$ is marked on the side $[AC]$ such that $\angle ABD = \angle DBC$. Knowing that $AB = 10$ and $CD = 3$, what is the length of the side $[BC]$?
A convex quadrilateral $ABCD$ is given where $\angle DAB =\angle ABC = 120^o$ and $CD = 3$,$BC = 2$, $AB = 1$. Calculate the length of segment $AD$.
The line joining the midpoints of two opposite sides of a convex quadrilateral makes equal angles with the diagonals. Show that the diagonals are equal.
In the triangle $ABC$ it is known that $2AC=AB$ and $\angle A = 2\angle B$. In this triangle draw the angle bisector $AL$, and mark point $M$, the midpoint of the side $AB$. It turned out that $CL = ML$. Prove that $\angle B= 30^o$.
(Hilko Danilo)
In a convex quadrilateral $ABCD$, $\angle ABC = 90^o$ , $\angle BAC = \angle CAD$, $AC = AD, DH$ is the alltitude of the triangle $ACD$. In what ratio does the line $BH$ divide the segment $CD$?
Let $\Gamma$ be a semicircle with diameter $AB$. On this diameter is selected a point $C$, and on the semicircle are selected points $D$ and $E$ so that $E$ lies between $B$ and $D$. It turned out that $\angle ACD = \angle ECB$. The intersection point of the tangents to $\Gamma$ at points $D$ and $E$ is denoted by $F$. Prove that $\angle EFD=\angle ACD+ \angle ECB$.
Let $\vartriangle ABC$ be a triangle. The angle bisector of $\angle CAB$ intersects$ BC$ at $L$. On the interior of line segments $AC$ and $AB$, two points, $M$ and $N$, respectively, are chosen in such a way that the lines $AL, BM$ and $CN$ are concurrent, and such that $\angle AMN = \angle ALB$. Prove that $\angle NML = 90^o$.
On the sides $(AB)$ and $(AC)$ of the triangle $ABC$ are considered the points $M$ and $N$ respectively so that $ \angle ABC =\angle ANM$. Point $D$ is symmetric of point $A$ with respect to $B$, and $P$ and $Q$ are the midpoints of the segments $[MN]$ and $[CD]$, respectively. Prove that the points $A, P$ and $Q$ are collinear if and only if $AC = AB \sqrt {2}$
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 $D$ be the point on side $AC$ of a triangle $ABC$ such that $BD$ bisects $\angle B$, and $E$ be the point on side $AB$ such that $3\angle ACE = 2\angle BCE$. Suppose that $BD$ and $CE$ intersect at a point $P$ with $ED = DC = CP$. Determine the angles of the triangle.
Let $\gamma$ be a semicircle with diameter $AB$ . A creek is built with origin in $A$ , which has its vertices alternately in the diameter $AB$ and in the semicircle $\gamma$ , so that its sides make equal angles $\alpha$ with the diameter (but alternately in either direction). It is requested:
a) Values of the angle $\alpha$ for the ravine to pass through the other end $B$ of the diameter.
b) The total length of the ravine, in the case that it ends in $B$ , as a function of the length $d$ of the diameter and of the angle $\alpha$ .
[img]https://cdn.artofproblemsolving.com/attachments/3/c/54c71cdf1bf8fbb3bdfa38f4b14a1dc961c5fe.png[/img]
Let $M$ be the midpoint of side $BC$ of an equilateral triangle $ABC$. The point $D$ is on $CA$ extended such that $A$ is between $D$ and $C$. The point $E$ is on $AB$ extended such that $B$ is between $A$ and $E$, and $|MD| = |ME|$. The point $F$ is the intersection of $MD$ and $AB$. Prove that $\angle BFM = \angle BME$.
Let $ABCD$ be a convex quadrilateral with $\angle BCD= 120^o, \angle {CBA} = 45^o, \angle {CBD} = 15^o$ and $\angle {CAB} = 90^o$. Show that $AB = AD$.
Five points $A, B, C, D$, and $E$ in three-dimensional Euclidean space have the property that $AB = BC = CD = DE = EA = 1$ and $\angle ABC = \angle BCD =\angle CDE = \angle DEA = 90^o$ . Find all possible $\cos(\angle EAB)$.
Let $ABC$ be a triangle with $\angle CAB = 2 \angle ABC$. Assume that a point $D$ is inside the triangle $ABC$ exists such that $AD = BD$ and $CD = AC$. Show that $\angle ACB = 3 \angle DCB$.
Let $O$ be the circumcenter of a triangle $ABC$. The projections of points $D$ and $X$ to the sidelines of the triangle lie on lines $\ell $ and $L $ such that $\ell // XO$. Prove that the angles formed by $L$ and by the diagonals of quadrilateral $ABCD$ are equal.
Let $D$ be a point inside an acute triangle $ABC$, such that $\angle BAD = \angle DBC$ and $\angle DAC = \angle BCD$. Let $P$ be a point on the circumcircle of the triangle $ADB$. Suppose $P$ are itself outside the triangle $ABC$. A line through $P$ intersects the ray $BA$ in $X$ and ray $CA$ in $Y$, so that $\angle XPB = \angle PDB$. Show that $BY$ and $CX$ intersect on $AD$.
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.
Let $\Omega$ be the circumcircle of an acute-angled triangle $ABC$. Let $D$ be the midpoint of the minor arc $AB$ of $\Omega$. A circle $\omega$ centered at $D$ is tangent to $AB$ 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
In an isosceles right triangle $ABC$ the right angle is at vertex $C$. On the side $AC$ points $K, L$ and on the side $BC$ points $M, N$ are chosen so that they divide the corresponding side into three equal segments. Prove that there is exactly one point $P$ inside the triangle $ABC$ such that $\angle KPL = \angle MPN = 45^o$.
Let $ABCDE$ be a convex pentagon such that $\angle BAC = \angle CAD = \angle DAE$ and $\angle ABC = \angle ACD = \angle ADE$. Diagonals $BD$ and $CE$ meet at $P$. Prove that $AP$ bisects side $CD$.
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 $ABCD$ be a convex quadrilateral with $\angle DAB =\angle ABC =\angle BCD > 90^o$. The circle circumscribed around the triangle $ABC$ intersects the sides $AD$ and $CD$ at points $K$ and $L$, respectively, different from any vertex of the quadrilateral $ABCD$ . Segments $AL$ and $CK$ intersect at point $P$. Prove that $\angle ADB =\angle PDC$.