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 $AC = 2 MD$.

Let $ABC$ be a triangle with $AB>AC$. Let $D$ be the foot of the internal angle bisector of $A$. Points $F$ and $E$ are on $AC,AB$ respectively such that $B,C,F,E$ are concyclic. Prove that the circumcentre of $DEF$ is the incentre of $ABC$ if and only if $BE+CF=BC$.

The bisector of angle $A$ in parallelogram $ABCD$ intersects side $BC$ at $M$ and the bisector of $\angle AMC$ passes through point $D$. Find angles of the parallelogram if it is known that $\angle MDC = 45^o$. [img]https://cdn.artofproblemsolving.com/attachments/e/7/7cfb22f0c26fe39aa3da3898e181ae013a0586.png[/img]

Let $ABC$ be a triangle and let $D$ be a point on $BC$ so that $AD$ bisects the angle $\angle BAC$. The common tangents of the circles $(BAD)$, $(CAD)$ meet at the point $A'$. The points $B'$, $C'$ are defined similarly. Show that $A'$, $B'$, $C'$ are collinear.

In the triangle $ABC$, the bisector of angle $A$ is equal to the half-sum of the height and median drawn from vertex $A$. Prove that if $\angle A$ is obtuse, then $AB = AC$.

In the convex quadrilateral $ABCD$ we have that $\angle BCD = \angle ADC \ge 90 ^o$. The bisectors of $\angle BAD$ and $\angle ABC$ intersect in $M$. Prove that if $M \in CD$, then $M$ is the middle of $CD$.

Two circles are tangent externaly at a point $B$. A line tangent to one of the circles at a point $A$ intersects the other circle at points $C$ and $D$. Show that $A$ is equidistant to the lines $BC$ and $BD$.

Let $ABC$ be an acute triangle with $| AB | > | AC |$. Let $D$ be the foot of the altitude from $A$ to $BC$, let $K$ be the intersection of $AD$ with the internal bisector of angle $B$, Let $M$ be the foot of the perpendicular from $B$ to $CK$ (it could be in the extension of segment $CK$) and$ N$ the intersection of $BM$ and $AK$ (it could be in the extension of the segments). Let $T$ be the intersection of$ AC$ with the line that passes through $N$ and parallel to $DM$. Prove that $BM$ is the internal bisector of the angle $\angle TBC$

A quadrilateral A$BCD$ is inscribed into a circle with center $O$. Points $C', D'$ are the reflections of the orthocenters of triangles $ABD$ and $ABC$ at point $O$. Lines $BD$ and $BD'$ are symmetric with respect to the bisector of angle $ABC$. Prove that lines $AC$ and $AC'$ are symmetric with respect to the bisector of angle $DAB$.

Let $AL$ and $BK$ be angle bisectors in the non-isosceles triangle $ABC$ ($L$ lies on the side $BC$, $K$ lies on the side $AC$). The perpendicular bisector of $BK$ intersects the line $AL$ at point $M$. Point $N$ lies on the line $BK$ such that $LN$ is parallel to $MK$. Prove that $LN = NA$.

Let $P$ be an interior point of an acute-angled triangle $ABC$. The line $BP$ meets the line $AC$ at $E$, and the line $CP$ meets the line $AB$ at $F$. The lines $AP$ and $EF$ intersect each other at $D$. Let $K$ be the foot of the perpendicular from the point $D$ to the line $BC$. Show that the line $KD$ bisects the angle $\angle EKF$.

Let $ ABC$ be a triangle. The $ A$-excircle of triangle $ ABC$ has center $ O_{a}$ and touches the side $ BC$ at the point $ A_{a}$. The $ B$-excircle of triangle $ ABC$ touches its sidelines $ AB$ and $ BC$ at the points $ C_{b}$ and $ A_{b}$. The $ C$-excircle of triangle $ ABC$ touches its sidelines $ BC$ and $ CA$ at the points $ A_{c}$ and $ B_{c}$. The lines $ C_{b}A_{b}$ and $ A_{c}B_{c}$ intersect each other at some point $ X$. Prove that the quadrilateral $ AO_{a}A_{a}X$ is a parallelogram. [i]Remark.[/i] The $ A$[i]-excircle[/i] of a triangle $ ABC$ is defined as the circle which touches the segment $ BC$ and the extensions of the segments $ CA$ and $ AB$ beyound the points $ C$ and $ B$, respectively. The center of this circle is the point of intersection of the interior angle bisector of the angle $ CAB$ and the exterior angle bisectors of the angles $ ABC$ and $ BCA$. Similarly, the $ B$-excircle and the $ C$-excircle of triangle $ ABC$ are defined. [hide="Source of the problem"][i]Source of the problem:[/i] Theorem (88) in: John Sturgeon Mackay, [i]The Triangle and its Six Scribed Circles[/i], Proceedings of the Edinburgh Mathematical Society 1 (1883), pages 4-128 and drawings at the end of the volume.[/hide]

In triangle $ABC$, bisectors are drawn $AA_1$ and $CC_1$. Prove that if the length of the perpendiculars drawn from the vertex $B$ on lines $AA1$ and $CC_1$ are equal, then $\vartriangle ABC$ is isosceles.

Let $ABC$ be a triangle with $AB < AC$. A point $P$ on the circumcircle of $ABC$ (on the same side of $BC$ as $A$) is chosen in such a way that $BP = CP$. Let $BP$ and the angle bisector of $\angle BAC$ intersect at $Q$, and let the line through $Q$ and parallel to $BC$ intersect $AC$ at $R$. Prove that $BR = CR$.

Given triangle $ABC$, $D$ is the foot of the external angle bisector of $A$, $I$ its incenter and $I_a$ its $A$-excenter. Perpendicular from $I$ to $DI_a$ intersects the circumcircle of triangle in $A'$. Define $B'$ and $C'$ similarly. Prove that $AA',BB'$ and $CC'$ are concurrent. [i]proposed by Amirhossein Zabeti[/i]