Found problems: 509
The sum of the angles on the bigger base of a trapezoid is $90^o$. Prove that the line segment whose ends are the midpoints of the bases, is equal to the line segment whose ends are the midpoints of the diagonals.
In $\vartriangle ABC$ with $AB > AC$, the tangent to the circumcircle at $A$ intersects line $BC$ at $P$. Let $Q$ be the point on $AB$ such that $AQ = AC$, and $A$ lies between $B$ and $Q$. Let $R$ be the point on ray $AP$ such that $AR = CP$. Let $X, Y$ be the midpoints of $AP, CQ$ respectively. Prove that $CR = 2XY$ .
Through the vertices $A, B$ of the parallelogram $ABCD$ passes a circle that intersects for the second time diagonals $BD$ and $AC$ at points $X$ and $Y$, respectively. The circumsccribed circle of $\vartriangle ADX$ intersects diagonal $AC$ for the second time at the point $Z$. Prove that $AY = CZ$.
In the triangle $ABC$, $\angle .ACB = 60^o$, and the bisectors $AA_1$ and $BB_1$ intersect at the point $M$. Prove that $MB_1 = MA_1$.
Let $ABCD$ be a convex quadrilateral and $K$ be the common point of rays $AB$ and $DC$. There exists a point $P$ on the bisectrix of angle $AKD$ such that lines $BP$ and $CP$ bisect segments $AC$ and $BD$ respectively. Prove that $AB = CD$.
Let $ABCD$ be a cyclic quadrilateral with $\angle ABC = 60^o$ and $| BC | = | CD |$. Prove that $|CD| + |DA| = |AB|$
In $\vartriangle ABC$ the incircle is tangent to $AB$ at $D$. Let $P$ be a point on $BC$ different from $B$ and $C$, and let $K$ and $L$ be incenters of $\vartriangle ABP$ and $\vartriangle ACP$ respectively. Suppose that the circumcircle of $\vartriangle KP L$ cuts $AP$ again at $Q$. Prove that $AD = AQ$.
Consider equilateral triangle $ABC$ and suppose that there exist three distinct points $X, Y,Z$ lie inside triangle $ABC$ such that
i) $AX = BY = CZ$
ii) The triplets of points $(A,X,Z), (B,Y,X), (C,Z,Y )$ are collinear in that order.
Prove that $XY Z$ is an equilateral triangle.
Let $ABC$ ($BC > AC$) be an acute triangle with circumcircle $k$ centered at $O$. The tangent to $k$ at $C$ intersects the line $AB$ at the point $D$. The circumcircles of triangles $BCD, OCD$ and $AOB$ intersect the ray $CA$ (beyond $A$) at the points $Q, P$ and $K$, respectively, such that $P \in (AK)$ and $K \in (PQ)$. The line $PD$ intersects the circumcircle of triangle $BKQ$ at the point $T$, so that $P$ and $T$ are in different halfplanes with respect to $BQ$. Prove that $TB = TQ$.
Consider the square $ABCD$ and the point $E$ inside the angle $CAB$, such that $\angle BAE =15^o$, and the lines $BE$ and $BD$ are perpendicular. Prove that $AE = BD$.
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)
Point $P$ is chosen inside triangle $ABC$ so that $\angle APC+\angle ABC=180^o$ and $BC=AP.$ On the side $AB$, a point $K$ is chosen such that $AK = KB + PC$. Prove that $CK \perp AB$.
Let $ABC$ be an acute triangle, and let $D$ be the foot of the altitude through $A$. On $AD$, there are distinct points $E$ and $F$ such that $|AE| = |BE|$ and $|AF| =|CF|$. A point$ T \ne D$ satis es $\angle BTE = \angle CTF = 90^o$. Show that $|TA|^2 =|TB| \cdot |TC|$.
Let $BH$ be an altitude of a right-angled triangle $ABC$ ($\angle B = 90^o$). The incircle of triangle $ABH$ touches $AB,AH$ in points $H_1, B_1$, the incircle of triangle $CBH$ touches $CB,CH$ in points $H_2, B_2$, point $O$ is the circumcenter of triangle $H_1BH_2$. Prove that $OB_1 = OB_2$.
An isosceles triangle $ABC$ with base $AC$ is given. On the rays $CA$, $AB$ and $BC$, the points $D, E$ and $F$ were marked, respectively, in such a way that $AD = AC$, $BE = BA$ and $CF = CB$. Find the sum of the angles $\angle ADB$, $\angle BEC$ and $\angle CFA$.
Different points $A_1,A_2,A_3,A_4,A_5$ lie on a circle so that $A_1A_2 = A_2A_3 = A_3A_4 =A_4A_5$. Let $A_6$ be the diametrically opposite point to $A_2$, and $A_7$ be the intersection of $A_1A_5$ and $A_3A_6$. Prove that the lines $A_1A_6$ and $A_4A_7$ are perpendicular
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 $ABC$ be a triangle inscribed in the circle $(O)$. Tangents at $B,C$ of the circles $(O)$ meet at $T$ . Let $M,N$ be the points on the rays $BT,CT$ respectively such that $BM = BC = CN$. The line through $M$ and $N$ intersects $CA,AB$ at $E, F$ respectively; $BE$ meets $CT$ at $P, CF$ intersects $BT$ at $Q$. Prove that $AP = AQ$.
Trần Quang Hùng
Point $P$ is chosen inside triangle $ABC$ so that $\angle APC+\angle ABC=180^o$ and $BC=AP.$ On the side $AB$, a point $K$ is chosen such that $AK = KB + PC$. Prove that $CK \perp AB$.
Through the midpoint $ S $ of the segment $ MN $ with endpoints lying on the legs of an isosceles triangle, a straight line is drawn parallel to the base of the triangle, intersecting its legs at points $ K $ and $ L $. Prove that the orthogonal projection of the segment $ MN $ onto the base of the triangle is equal to the segment $ KL $.
Let $ABC$ be a triangle with $AC> AB$. Let $P$ be the intersection of $BC$ and the tangent through $A$ around the triangle $ABC$. Let $Q$ be the point on the straight line $AC$, so that $AQ = AB$ and $A$ is between $C$ and $Q$. Let $X$ and $Y$ be the center of $BQ$ and $AP$. Let $R$ be the point on $AP$ so that $AR = BP$ and $R$ is between $A$ and $P$. Show that $BR = 2XY$.
The convex pentagon $ABCDE$ is cyclic and $AB = BD$. Let point $P$ be the intersection of the diagonals $AC$ and $BE$. Let the straight lines $BC$ and $DE$ intersect at point $Q$. Prove that the straight line $PQ$ is parallel to the diagonal $AD$.
In the triangle $ABC$, let $P$, $Q$, and $R$ lie on the sides $BC$, $AC$, and $AB$ respectively, such that $AQ = AR$, $BP = BR$ and $CP = CQ$. Let $\angle PQR=75^o$ and $\angle PRQ=35^o$. Calculate the measures of the angles of the triangle $ABC$.
In a right triangle $ABC$ with $\angle BAC =90^o $and $\angle ABC= 54^o$, point $M$ is the midpoint of the hypotenuse $[BC]$ , point $D$ is the foot of the angle bisector drawn from the vertex $C$ and $AM \cap CD = \{E\}$. Prove that $AB= CE$.
In an acute triangle $\vartriangle ABC$, $D$ is the foot of altitude from $A$ to $BC$. Suppose that $AD = CD$, and define $N$ as the intersection of the median $CM$ and the line $AD$. Prove that $\vartriangle ABC$ is isosceles if and only if $CN = 2AM$.