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 $ABC$ 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 $BD=2CD$.

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.

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 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 $\angle NA'T = \angle ADT$.

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 $MNK$.

Let $AD$ and $AE$ be the altitude and median of triangle $ABC$, in with $\angle B = 2\angle C$. Prove that $AB = 2DE$.

Let $ABC$ be atriangle with $\overline{AC}> \overline{BC}$ and incircle $k$. Let $M,W,L$ be the intersections of the median, angle bisector and altitude from point $C$ respectively. The tangent to $k$ passing through $M$, that is different from $AB$, touch $k$ in $T$. Prove that the angles $\angle MTW$ and $\angle TLM$ are equal.

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 $\angle PDA = \angle AED$.

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 $\angle CPM = \angle DPL$.

Given the rectangle $ABCD$. The points $E ,F$ lie on the segments $(BC) , (DC)$ respectively, such that $\angle DAF = \angle FAE$. Proce that if $DF + BE = AE$ then $ABCD$ is square.

Let $O$ be the center of the circle of the triangle $ABC$ with $AB \ne AC$. Furthermore, let $S$ and $T$ be points on the rays $AB$ and $AC$, such that $\angle ASO = \angle ACO$ and $\angle ATO = \angle ABO$. Show that $ST$ bisects the segment $BC$.

Let $ABCD$ be an inscribed quadrilateral. Denote the midpoints of the sides $AB, BC, CD$ and $DA$ through $M, L, N$ and $K$, respectively. It turned out that $\angle BM N = \angle MNC$. Prove that: i) $\angle DKL = \angle CLK$. ii) in the quadrilateral $ABCD$ there is a pair of parallel sides.

Given a quadrilateral $ABCD$ with $\angle ABC = \angle ADC$. Let $BM$ be the altitude of the triangle $ABC$, and $M$ belongs to $AC$. Point $M'$ is marked on the diagonal $AC$ so that $$\frac{AM \cdot CM'}{ AM' \cdot CM}= \frac{AB \cdot CD }{ BC \cdot AD}$$ Prove that the intersection point of $DM'$ and $BM$ coincides with the orthocenter of the triangle $ABC$. (M. Zhikhovich)

Each pair of opposite sides of a convex hexagon has the following property: the distance between their midpoints is equal to $\dfrac{\sqrt{3}}{2}$ times the sum of their lengths. Prove that all the angles of the hexagon are equal.

Let $ABCD$ be a rectangle. The point $E$ lies on side $ \overline{AB}$ and the point $F$ is lies side $ \overline{AD}$, such that $\angle FEC=\angle CEB$ and $\angle DFC=\angle CFE$. Determine the measure of the angle $\angle FCE$ and the ratio $AD/AB$.

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 $ PQ $ be a chord cut by the line $ AB $ on circle $ S_3 $, and $ K $ be the midpoint of $ CD $. Prove that $ \angle PKC = \angle QKC $.

In triangle $ABC$, $|AB| <|BC|$ holds. Point $I$ is the center of the circle inscribed in that triangle. Let $M$ be the midpoint of the side $AC$, and $N$ be the midpoint of the arc $AC$ of the circumcircle of that triangle containing point $B$. Prove that $\angle IMA = \angle INB$.

A (convex) trapezoid $ABCD$ is good, if it is inscribed in a circle, sides $AB$ and $CD$ are the bases and $CD$ is shorter than $AB$. For a good trapezoid $ABCD$ the following terms are defined: $\bullet$ The parallel to $AD$ passing through $B$ intersects the extension of side $CD$ at point $S$. $\bullet$ The two tangents passing through $S$ on the circumircle of the trapezoid touch the circle at $E$ and $F$, where $E$ lies on the same side of the straight line $CD$ as $A$. Give the simplest possible equivalent condition (expressed in side lengths and / or angles of the trapezoid) so that with a good trapezoid $ABCD$ the two angles $\angle BSE$ and $\angle FSC$ have the same measure. (Walther Janous)

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.

Diagonals of an isosceles trapezoid $ABCD$ with bases $BC$ and $AD$ are perpendicular. Let $DE$ be the perpendicular from $D$ to $AB$, and let $CF$ be the perpendicular from $C$ to $DE$. Prove that angle $DBF$ is equal to half of angle $FCD$.

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]

Given a convex pentagon $ABCDE$ in which $\angle ABC = \angle AED = 90^o$, $\angle BAC= \angle DAE$. Let $K$ be the midpoint of the side $CD$, and $P$ the intersection point of lines $AD$ and $BK$, $Q$ be the intersection point of lines $AC$ and $EK$. Prove that $BQ = PE$.

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 $\angle AMD = \angle CMD$.

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 $\angle IMA = \angle INB$.