Found problems: 361
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.
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 $ K $ be the midpoint of the chord cut by the line $ AB $ on circles $ S_3 $. Prove that $ \angle CKA = \angle DKA $.
Triangles $ABC$ and $DEF$ are such that $\angle A = \angle D, AB = DE = 17, BC = EF = 10$ and $AC - DF = 12$. What is $AC + DF$?
Given an isosceles triangle $ABC$ such that $|AB|=|AC|$ . Let $M$ be the midpoint of the segment $BC$ and let $P$ be a point other than $A$ such that $PA\parallel BC$. The points $X$ and $Y$ are located respectively on rays $PB$ and $PC$, so that the point $B$ is between $P$ and $X$, the point $C$ is between $P$ and $Y$ and $\angle PXM=\angle PYM$. Prove that the points $A,P,X$ and $Y$ are concyclic.
In a triangle $ABC$, $T$ is the centroid and $ \angle TAB = \angle ACT$. Find the maximum possible value of $sin \angle CAT +sin \angle CBT$.
The diagonals of quadrilateral $ABCD$ intersect in point $E$. Given that $|AB| =|CE|$, $|BE| = |AD|$, and $\angle AED = \angle BAD$, determine the ratio $|BC|:|AD|$.
Let $ABC$ be a triangle with $AH$ altitude. The point $K$ is chosen on the segment $AH$ as follows such that $AH =3KH$. Let $O$ be the center of the circle circumscribed around by triangle $ABC, M$ and $N$ be the midpoints of $AC$ and AB respectively. Lines $KO$ and $MN$ intersect at the point $Z$, a perpendicular to $OK$ passing through point $Z$ intersects lines $AC$ and $AB$ at points $X$ and $Y$ respectively. Prove that $\angle XKY =\angle CKB$.
Let $ABC$ be a triangle in which $\angle ABC = 45^o$ and $\angle BAC > 90^o$. Let $O$ be the midpoint of the side $[BC]$. Consider the point $M \in (AC)$ such that $\angle COM =\angle CAB$. Perpendicular from $M$ on $AC$ intersects line $AB$ at point $P$.
a) Find the measure of the angle $\angle BCP$.
b) Show that if $\angle BAC = 105^o$, then $PB = 2MO$.
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 $\angle BAP$.
Let $ABC$ be an acute triangle and let $P,Q$ be points on $BC$ such that $\angle QAC =\angle ABC$ and $\angle PAB = \angle ACB$. We extend $AP$ to $M$ so that $ P$ is the midpoint of $AM$ and we extend $AQ$ to $N$ so that $Q$ is the midpoint of $AN$. If T is the intersection point of $BM$ and $CN$, show that quadrilateral $ABTC$ is cyclic.
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.
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)
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 $BD + BA = AC$.
Let $ABC$ be an acute triangle. Let $P$ be a point such that $PB$ and $PC$ are tangent to circumcircle of $ABC$. Let $X$ and $Y$ be variable points on $AB$ and $AC$, respectively, such that $\angle XPY = 2\angle BAC$ and $P$ lies in the interior of triangle $AXY$. Let $Z$ be the reflection of $A$ across $XY$. Prove that the circumcircle of $XYZ$ passes through a fixed point.
(Dominik Burek, Poland)
Let $ABC$ be a triangle with $AB > AC$. Let $D$ be a point on the side $AB$ such that $DB = DC$ and let $M$ be the midpoint of $AC$. The line parallel to $BC$ passing through $D$ intersects the line $BM$ in $K$. Show that $\angle KCD = \angle DAC.$
Consider $D$ the midpoint of the base $[BC]$ of the isosceles triangle ABC in which $\angle BAC < 90^o$. On the perpendicular from $B$ on the line $BC$ consider the point $E$ such that $\angle EAB= \angle BAC$, and on the line passing though $C$ parallel to the line $AB$ we consider the point $F$ such that $F$ and $D$ are on different side of the line $AC$ and $\angle FAC = \angle CAD$. Prove that $AE = CF$ and $BF = EF$
The point $ M$ is chosen inside parallelogram $ ABCD$. Show that $ \angle MAB$ is congruent to $ \angle MCB$, if and only if $ \angle MBA$ and $ \angle MDA$ are congruent.
Let $ABC$ be a triangle with $AB=AC$. Let $D$ be a point on $BC$, and $E$ a point on $AD$ such that $\angle BED=\angle BAC=2\angle CED$. Prove that $BD=2CD$.
In a triangle ABC we have $\angle C = \angle A + 90^o$. The point $D$ on the continuation of $BC$ is given such that $AC = AD$. A point $E$ in the side of $BC$ in which $A$ doesn’t lie is chosen such that $\angle EBC = \angle A, \angle EDC = \frac{1}{2} \angle A$ . Prove that $\angle CED = \angle ABC$.
by Morteza Saghafian
Let $ABCD$ be a trapezium with $AD // BC$. Suppose $K$ and $L$ are, respectively, points on the sides $AB$ and $CD$ such that $\angle BAL = \angle CDK$. Prove that $\angle BLA = \angle CKD$.
Consider a triangle $ABC$ with $\angle ACB = 2 \angle CAB $ and $\angle ABC> 90 ^ \circ$. Consider the perpendicular on $AC$ that passes through $A$ and intersects $BC$ at $D$, prove that $$\frac {1} {BC} - \frac {2} {DC} = \frac {1} {CA} $$
Let $ABC$ be an acute triangle, where $AB$ is the smallest side and let $D$ be the midpoint of $AB$. Let $P$ be a point in the interior of the triangle $ABC$ such that $\angle CAP = \angle CBP = \angle ACB$. From the point $P$, we draw perpendicular lines on $BC$ and $AC$ where the intersection point with $BC$ is $M$, and with $AC$ is $N$ . Through the point $M$ we draw a line parallel to $AC$, and through $N$ parallel to $BC$. These lines intercept at the point $K$. Prove that $D$ is the center of the circumscribed circle for the triangle $MNK$.
Let $\vartriangle ABC$ be a triangle with circumcenter $O$, orthocenter $H$, and circumcircle $\omega$. $AA'$, $BB'$ and $CC'$ are altitudes of $\vartriangle ABC$ with $A'$ in $BC$, $B'$ in $AC$ and $C'$ in $AB$. $P$ is a point on the segment $AA'$. The perpenicular line to $B'C'$ from $P$ intersects $BC$ at $K$. $AA'$ intersects $\omega$ at $M \ne A$. The lines $MK$ and $AO$ intersect at $Q$. Prove that $\angle CBQ = \angle PBA$.
In the trapezoid $ABCD$ , the base $AB$ is smaller than the $CD$ base. The point $K$ is chosen such that $AK$ is parallel to BC and $BK$ is parallel to $AD$. The points $P$ and $Q$ are chosen on the $AK$ and $BK$ rays respectively, such that $\angle ADP = \angle BCK$ and $\angle BCQ = \angle ADK$.
(a) Show that the lines $AD, BC$ and $PQ$ go through the same point.
(b) Assuming that the circumscribed circumferences of the $APD$ and $BCQ$ triangles intersect at two points, show that one of those points belongs to the line $PQ$.
In the triangle $ABC$ on the ray $BA$ mark the point $K$ so that $\angle BCA= \angle KCA$ , and on the median $BM$ mark the point $T$ so that $\angle CTK=90^o$ . Prove that $\angle MTC=\angle MCB$ .