Found problems: 361
Points $P$ and $Q$ lies inside triangle $ABC$ such that $\angle ACP =\angle BCQ$ and $\angle CAP = \angle BAQ$. Denote by $D,E$, and $F$ the feet of perpendiculars from $P$ to lines $BC,CA$, and $AB$, respectively. Prove that if $\angle DEF = 90^o$, then $Q$ is the orthocenter of triangle $BDF$.
Given a square $ABCD$. Find the locus of points $M$ such that $\angle AMB = \angle CMD$.
Let $ABC$ be a triangle with $AB > AC$. Point $P \in (AB)$ is such that $\angle ACP = \angle ABC$. Let $D$ be the reflection of $P$ into the line $AC$ and let $E$ be the point in which the circumcircle of $BCD$ meets again the line $AC$. Prove that $AE = AC$.
On the sides $AC$ and $AB$ of the triangle $ABC$, the points $D$ and $E$ were chosen such that $\angle ABD =\angle CBD$ and $3 \angle ACE = 2\angle BCE$. Let $H$ be the point of intersection of $BD$ and $CE$, and $CD = DE = CH$. Find the angles of triangle $ABC$.
Prove that in all triangles $\vartriangle ABC$ with $\angle A = 2 \angle B$ it holds that, if $D$ is the foot of the perpendicular from $C$ to the perpendicular bisector of $AB$, $\frac{AC}{DC}$ is constant for any value of $\angle B$.
In triangle $ABC$, point $M$ is the midpoint of $BC$, $P$ the point of intersection of the tangents at points $B$ and $C$ of the circumscribed circle of $ABC$, $N$ is the midpoint of the segment $MP$. The segment $AN$ meets the circumcircle $ABC$ at the point $Q$. Prove that $\angle PMQ = \angle MAQ$.
The points $A, B, C, D$ lie in that order on a circle. The segments $AC$ and $BD$ intersect at the point $P$. The point $B'$ lies on the line $AB$ such that $A$ is between $B$ and $B'$ and $|AB'| = |DP |$. The point $C'$ lies on the line $CD$ such that $D$ is between $C$ and $C'$ lies and $|DC' | = |AP|$. Prove that $\angle B'PC' = \angle ABD'$.
[img]https://cdn.artofproblemsolving.com/attachments/2/2/7ec65246ff5ecfebc25ca13f3709d1791ceb6c.png[/img]
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Circles $\omega_1$ and $\omega_2$ inscribed into equal angles $X_1OY$ and $Y OX_2$ touch lines $OX_1$ and $OX_2$ at points $A_1$ and $A_2$ respectively. Also they touch $OY$ at points $B_1$ and $B_2$. Let $C_1$ be the second common point of $A_1B_2$ and $\omega_1, C_2$ be the second common point of $A_2B_1$ and $\omega_2$. Prove that $C_1C_2$ is the common tangent of two circles.
Let $ABC$ be an isosceles triangle with $AB = AC$. The points $D, E$ and $F$ are taken on the sides $BC, CA$ and $AB$, respectively, so that $\angle F DE = \angle ABC$ and $FE$ is not parallel to $BC$. Prove that the line $BC$ is tangent to the circumcircle of $\vartriangle DEF$ if and only if $D$ is the midpoint of the side $BC$.
On the plane are given four distinct points $A,B,C,D$ on a line $g$ in this order, at the mutual distances $AB = a, BC = b, CD = c$.
(a) Construct (if possible) a point $P$ outside line $g$ such that $\angle APB =\angle BPC =\angle CPD$.
(b) Prove that such a point $P$ exists if and only if $ (a+b)(b+c) < 4ac$
Let $ABC$ be an acute triangle, and let $D$ be the foot of the altitude from $A$. The circle with centre $A$ passing through $D$ intersects the circumcircle of triangle $ABC$ in $X$ and $Y$ , in such a way that the order of the points on this circumcircle is: $A, X, B, C, Y$ . Show that $\angle BXD = \angle CYD$.
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 $\angle ACB = 3 \angle DCB$.
As shown in the figure, given $\vartriangle ABC$ with $AB \perp AC$, $AB=BC$, $D$ is the midpoint of the side $AB$, $DF\perp DE$, $DE=DF$ and $BE \perp EC$. Prove that $\angle AFD= \angle CEF$.
[img]https://cdn.artofproblemsolving.com/attachments/9/2/f16a8c8c463874f3ccb333d91cdef913c34189.png[/img]
$ABCD$ is a parallelogram. $P$ is a point outside the parallelogram such that angles $\angle PAB$ and $\angle PCB$ have the same value but opposite orientation. Show that $\angle APB = \angle DPC$.
Inside the inscribed quadrilateral $ ABCD $, a point $ P $ is marked such that $ \angle PBC = \angle PDA $, $ \angle PCB = \angle PAD $. Prove that there exists a circle that touches the straight lines $ AB $ and $ CD $, as well as the circles circumscribed by the triangles $ ABP $ and $ CDP $.
Let $ABC$ be an acute triangle such that $AB$ is the shortest side of the triangle. Let $D$ be the midpoint of the side $AB$ and $P$ be an interior point of the triangle such that $\angle CAP = \angle CBP = \angle ACB$. Denote by M and $N$ the feet of the perpendiculars from $P$ to $BC$ and $AC$, respectively. Let $p$ be the line through $ M$ parallel to $AC$ and $q$ be the line through $N$ parallel to $BC$. If $p$ and $q$ intersect at $K$ prove that $D$ is the circumcenter of triangle $MNK$.
In triangle $ABC$ point $M$ is the midpoint of side $AB$, and point $D$ is the foot of altitude $CD$.
Prove that $\angle A = 2\angle B$ if and only if $AC = 2MD$
The median $AM$ is drawn in the triangle $ABC$ ($AB \ne AC$). The point $P$ is the foot of the perpendicular drawn on the segment $AM$ from the point $B$. On the segment $AM$ we chose such a point $Q$ that $AQ = 2PM$. Prove that $\angle CQM = \angle BAM$.
In the cyclic quadrilateral $ABCD$, the diagonals $AC$ and $BD$ intersect at $P$. Let $O$ be the center of the circumcircle $ABCD$, and $E$ a point of the extension of $OC$ beyond $C$. A parallel line to $CD$ is drawn through $E$ that cuts the extension of $OD$ beyonf $D$ at $F$. Let $Q$ be a point interior to $ABCD$, such that $\angle AFQ = \angle BEQ$ and $\angle FAQ = \angle EBQ$. Prove that $PQ \perp CD$.
Let the interior point $P$ of the convex quadrilateral $ABCD$ be such that $$|\angle PAD| = |\angle ADP| = |\angle CBP| = |\angle PCB| = |\angle CPD|.$$ Let $O$ be the center of the circumcircle of the triangle $CPD$. Prove that $|OA| = |OB|$.
A point$ O$ inside a parallelogram $ABCD$ satisfies $\angle AOB + \angle COD = \pi$. Prove that $\angle CBO = \angle CDO$.
Consider the pentagon $ABCDE$ such that $AB = AE = x$, $AC = AD = y$, $\angle BAE = 90^o$ and $\angle ACB = \angle ADE = 135^o$. It is known that $C$ and $D$ are inside the triangle $BAE$. Determine the length of $CD$ in terms of $x$ and $y$.
As shown in figure , points $D,E,F$ lies the sides $AB$, $BC$ , $CA$ of the acute angle $\vartriangle ABC$ respectively. If $\angle EDC = \angle CDF$, $\angle FEA=\angle AED$, $\angle DFB =\angle BFE$, prove that the $CD$, $AE$, $BF$ are the altitudes of $\vartriangle ABC$.
[img]https://cdn.artofproblemsolving.com/attachments/3/d/5ddf48e298ad1b75691c13935102b26abe73c1.png[/img]
Let $E$ be the midpoint of the side $CD$ of a square $ABCD$. Consider the point $M$ inside the square such that $\angle MAB = \angle MBC = \angle BME = x$. Find the angle $x$.
Let $ABC$ be a right angled triangle with $\angle A = 90^o$ and $AD$ its altitude. We draw parallel lines from $D$ to the vertical sides of the triangle and we call $E, Z$ their points of intersection with $AB$ and $AC$ respectively. The parallel line from $C$ to $EZ$ intersects the line $AB$ at the point $N$. Let $A' $ be the symmetric of $A$ with respect to the line $EZ$ and $I, K$ the projections of $A'$ onto $AB$ and $AC$ respectively. If $T$ is the point of intersection of the lines $IK$ and $DE$, prove that $\angle NA'T = \angle ADT$.