Found problems: 361
In the tetrahedron $DABC, AB=BC, \angle DBC =\angle DBA$. Prove that $AC \perp DB$.
In a right triangle $ABC $ with a right angle $\angle C $, n the sides $AC$ and $AB$, the points $M$ and $N$ are selected, respectively, that $CM = MN$ and $\angle MNB = \angle CBM$. Let the point $K$ be the projection of the point $C $ on the segment $MB $. Prove that the line $NK$ passes through the midpoint of the segment $BC$.
(Alex Klurman)
In square $ABCD$, $\overline{AC}$ and $\overline{BD}$ meet at point $E$.
Point $F$ is on $\overline{CD}$ and $\angle CAF = \angle FAD$.
If $\overline{AF}$ meets $\overline{ED}$ at point $G$, and if $\overline{EG} = 24$ cm, then find the length of $\overline{CF}$.
Let $\vartriangle ABC$ be a triangle such that angle $\angle ABC$ is less than angle $\angle ACB$. The bisector of angle $\angle BAC$ cuts side $BC$ at $D$. Let $E$ be on side $AB$ such that $\angle EDB = 90^o$ and $F$ on side $AC$ such that $\angle BED = \angle DEF$. Prove that $\angle BAD = \angle FDC$.
In a quadrilateral $ABCD$ we have $\angle A = \angle C = 90^o$. Let $E$ be a point in the interior of $ABCD$. Let $M$ be the midpoint of $BE$. Prove that $\angle ADB = \angle EDC$ if and only if $|MA| = |MC|$.
Let $\Gamma$ be a circle and $P$ be a point outside, $PA$ and $PB$ be tangents to $\Gamma$ , $A, B \in \Gamma$ . Point $K$ is an arbitrary point on the segment $AB$. The circumscirbed circle of $\vartriangle PKB$ intersects $\Gamma$ for the second time at point $T$, point $P'$ is symmetric to point $P$ wrt point $A$. Prove that $\angle PBT = \angle P'KA$.
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 $\angle BXD - \angle BYD = \angle BSD$.
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$.
Points $E$ and $F$ are the midpoints of sides $BC$ and $CD$ of square $ABCD$, respectively. Lines $AE$ and $BF$ meet at point $P$. Prove that $\angle PDA = \angle AED$.
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 $\angle BAP$.
Find the greatest $n$ for which there exist $n$ lines in space, passing through a single point, such that any two of them form the same angle.
Let $P$ be an interior point of the parallelogram $ABCD$. Prove that $\angle APB+ \angle CPD = 180^\circ$ if and only if $\angle PDC = \angle PBC$.
Let $ABC$ be a triangle with $\angle BAC = 90^o$. Angle bisector of the $\angle CBA$ intersects the segment $(AB)$ at point $E$. If there exists $D \in (CE)$ so that $\angle DAC = \angle BDE =x^o$ , calculate $x$.
Let $I$ be the incenter of the triangle $ABC$ ($AB < BC$). Let $M$ be the midpoint of $AC$, and let $N$ be the midpoint of the arc $AC$ of the circumcircle of $ABC$ which contains $B$. Prove that $\angle IMA = \angle INB$.
Inside the triangle $ABC$ there is a point $P$ such that $$\angle PAB=\angle PBC =\angle PCA = \phi.$$
Prove that $$\frac{1}{\sin^2 \phi}=\frac{1}{\sin^2 A} +\frac{1}{\sin^2 B} +\frac{1}{\sin^2 C}$$
The angle $POQ$ is given ($OP$ and $OQ$ are rays). Let $M$ and $N$ be points inside the angle $POQ$ such that $\angle POM = \angle QON$ and $\angle POM < \angle PON$. Consider two circles: one touches the rays $OP$ and $ON$, the other touches the rays $OM$ and $OQ$. Denote by $B$ and $C$ the points of their intersection. Prove that $\angle POC = \angle QOB$.
As shown in figure , $\odot O$ is the inscribed circle of trapezoid $ABCD$, and the tangent points are $E, F, G, H$, $AB \parallel CD$. The line passing through$ B$, parallel to $AD$ intersects extension of $DC$ at point $P$. The extension of $AO$ intersects $CP$ at point $Q$. If $AE=BE$ , prove that $\angle CBQ = \angle PBQ$.
[img]https://cdn.artofproblemsolving.com/attachments/d/2/7c3a04bb1c59bc6d448204fd78f553ea53cb9e.png[/img]
The quadrangle $ABCD$ is inscribed in a circle whose center $O$ lies inside it.
Prove that if $\angle BAO = \angle DAC$, then the diagonals of the quadrilateral are perpendicular.
Given the isosceles triangle $ABC$, where $AB = AC$. Let $D$ be a point in the segment $BC$ so that $BD = 2DC$. Suppose also that point $P$ lies on the segment $AD$ such that: $\angle BAC = \angle BP D$. Prove that $\angle BAC = 2\angle DP C$.
$ABC$ is a triangle and $D$ is an internal point such that $\angle DAB=\angle DBC =\angle DCA$. $O_a$ is the circumcenter of $DBC$. $O_b$ is the circumcenter of $DAC$. $O_c$ is the circumcenter of $DAB$. Show that if the area of $ABC$ and $O_aO_bO_c$ are equal then $ABC$ is equilateral.
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
In the quadrilateral $ABCD$, angles $B$ and $C$ are equal to $120^o$, $AB = CD = 1$, $CB = 4$. Find the length $AD$.
In the hexagon $ABCDEF$, all angles are equally large. The side lengths satisfy $AB = CD = EF = 3$ and $BC = DE = F A = 2$. The diagonals $AD$ and $CF$ intersect each other in the point $G$. The point $H$ lies on the side $CD$ so that $DH = 1$. Prove that triangle $EGH$ is equilateral.
Let $ABCD$ be rhombus $ABCD$ where the triangles $ABD$ and $BCD$ are equilateral. Let $M$ and $N$ be points on the sides $BC$ and $CD$, respectively, such that $\angle MAN = 30^o$. Let $X$ be the intersection point of the diagonals $AC$ and $BD$. Prove that $\angle XMN = \angle\ DAM$ and $\angle XNM = \angle BAN$.
Points $K$ and $L$ are taken on the sides $BC$ and $CD$ of a square $ABCD$ so that $\angle AKB = \angle AKL$. Find $\angle KAL$.