In triangle $ABC$, let $P$ and $R$ be the feet of the perpendiculars from $A$ onto the external and internal bisectors of $\angle ABC$, respectively; and let $Q$ and $S$ be the feet of the perpendiculars from $A$ onto the internal and external bisectors of $\angle ACB$, respectively. If $PQ = 7, QR = 6$ and $RS = 8$, what is the area of triangle $ABC$?

Let $ABC$ be a triangle, let $D$ be the touch point of the side $BC$ and the incircle of the triangle $ABC$, and let $J_b$ and $J_c$ be the incentres of the triangles $ABD$ and $ACD$, respectively. Prove that the circumcentre of the triangle $AJ_bJ_c$ lies on the bisector of the angle $BAC$.

Triangle $ABC$ has a right angle at $B$. Point $D$ lies on side $BC$ such that $3\angle BAD = \angle BAC$. Given $AC=2$ and $CD=1$, compute $BD$.

$ABC$ is a triangle such that $BC = 10$, $CA = 12$. Let $M$ be the midpoint of side $AC$. Given that $BM$ is parallel to the external bisector of $\angle A$, find area of triangle $ABC$. (Lines $AB$ and $AC$ form two angles, one of which is $\angle BAC$. The external angle bisector of $\angle A$ is the line that bisects the other angle.

Let $ABC$ be a triangle such that $|AB|=8$ and $|AC|=2|BC|$. What is the largest value of altitude from side $[AB]$? $ \textbf{(A)}\ 3\sqrt 2 \qquad\textbf{(B)}\ 3\sqrt 3 \qquad\textbf{(C)}\ 5 \qquad\textbf{(D)}\ \dfrac {16}3 \qquad\textbf{(E)}\ 6 $

In triangle $\triangle ABC$ , $I$ is the incenter and $M$ is the midpoint of arc $(BC)$ in the circumcircle of $(ABC)$not containing $A$. Let $X$ be an arbitrary point on the external angle bisector of $A$. Let $BX \cap (BIC) = T$. $Y$ lies on $(AXC)$ , different from $A$ , st $MA=MY$ . Prove that $TC || AY$ (Assume that $X$ is not on $(ABC)$ or $BC$)

Let $ ABCD$ be an isosceles trapezoid with $ AD \parallel BC$. Its diagonals $ AC$ and $ BD$ intersect at point $ M$. Points $ X$ and $ Y$ on the segment $ AB$ are such that $ AX \equal{} AM$, $ BY \equal{} BM$. Let $ Z$ be the midpoint of $ XY$ and $ N$ is the point of intersection of the segments $ XD$ and $ YC$. Prove that the line $ ZN$ is parallel to the bases of the trapezoid. [i]Author: A. Akopyan, A. Myakishev[/i]

Let $ABCD$ be a cyclic quadrilateral wih both diagonals perpendicular to each other and intersecting at point $O$. Let $E,F,G,H$ be the orthogonal projections of $O$ on sides $AB,BC,CD,DA$ respectively. a. Prove that $\angle EFG + \angle GHE = 180^o$ b. Prove that $OE$ bisects angle $\angle FEH$ .

Let $ABC$ be a triangle with $\angle{A} = 60^\circ$. The midperpendicular of segment $AB$ meets line $AC$ at point $C_1$. The midperpendicular of segment $AC$ meets line $AB$ at point $B_1$. Prove that line $B_1C_1$ touches the incircle of triangle $ABC$.

Let $ABCDEF$ be a convex hexagon such that $\angle A = \angle C = \angle E$ and $\angle B = \angle D = \angle F$ and the (interior) angle bisectors of $\angle A, ~\angle C,$ and $\angle E$ are concurrent. Prove that the (interior) angle bisectors of $\angle B, ~\angle D, $ and $\angle F$ must also be concurrent. [i]Note that $\angle A = \angle FAB$. The other interior angles of the hexagon are similarly described.[/i]

Let $D$ and $E$ be points on side $[AB]$ of a right triangle with $m(\widehat{C})=90^\circ$ such that $|AD|=|AC|$ and $|BE|=|BC|$. Let $F$ be the second intersection point of the circumcircles of triangles $AEC$ and $BDC$. If $|CF|=2$, what is $|ED|$? $ \textbf{(A)}\ \sqrt 2 \qquad\textbf{(B)}\ 1+\sqrt 2 \qquad\textbf{(C)}\ 2 \qquad\textbf{(D)}\ 2\sqrt 2 \qquad\textbf{(E)}\ \text{None of above} $

Let $I$ be the incenter of the scalene $\Delta ABC$, such, $AB<AC$, and let $I'$ be the reflection of point $I$ in line $BC$. The angle bisector $AI$ meets $BC$ at $D$ and circumcircle of $\Delta ABC$ at $E$. The line $EI'$ meets the circumcircle at $F$. Prove, that, $\text{(i) } \frac{AI}{IE}=\frac{ID}{DE}$ $\text{(ii) } IA=IF$

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 $AB = AC$.

A circle centered at $ O$ has radius $ 1$ and contains the point $ A$. Segment $ AB$ is tangent to the circle at $ A$ and $ \angle{AOB} \equal{} \theta$. If point $ C$ lies on $ \overline{OA}$ and $ \overline{BC}$ bisects $ \angle{ABO}$, then $ OC \equal{}$ [asy]import olympiad; unitsize(2cm); defaultpen(fontsize(8pt)+linewidth(.8pt)); labelmargin=0.2; dotfactor=3; pair O=(0,0); pair A=(1,0); pair B=(1,1.5); pair D=bisectorpoint(A,B,O); pair C=extension(B,D,O,A); draw(Circle(O,1)); draw(O--A--B--cycle); draw(B--C); label("$O$",O,SW); dot(O); label("$\theta$",(0.1,0.05),ENE); dot(C); label("$C$",C,S); dot(A); label("$A$",A,E); dot(B); label("$B$",B,E);[/asy] $ \textbf{(A)}\ \sec^2\theta \minus{} \tan\theta \qquad \textbf{(B)}\ \frac {1}{2} \qquad \textbf{(C)}\ \frac {\cos^2\theta}{1 \plus{} \sin\theta} \qquad \textbf{(D)}\ \frac {1}{1 \plus{} \sin\theta} \qquad \textbf{(E)}\ \frac {\sin\theta}{\cos^2\theta}$

Let $ABC$ be a triangle and a circle $\Gamma'$ be drawn lying outside the triangle, touching its incircle $\Gamma$ externally, and also the two sides $AB$ and $AC$. Show that the ratio of the radii of the circles $\Gamma'$ and $\Gamma$ is equal to $\tan^ 2 { \left( \dfrac{ \pi - A }{4} \right) }.$

Let $O$ be a point inside an acute angle with the vertex $A$ and $H, N$ be the feet of the perpendiculars drawn from $O$ onto the sides of the angle. Let point $B$ belong to the bisector of the angle, $K$ be the foot of the perpendicular from $B$ onto either side of the angle. Denote by $P,F$ the midpoints of the segments $AK,HN$ respectively. Known that $ON + OH = BK$, prove that $PF$ is perpendicular to $AB$. Ya. Konstantinovski

Let $D$ be a point on side $[BC]$ of triangle $ABC$ such that $[AD]$ is an angle bisector, $|BD|=4$, and $|DC|=3$. Let $E$ be a point on side $[AB]$ and different than $A$ such that $m(\widehat{BED})=m(\widehat{DEC})$. If the perpendicular bisector of segment $[AE]$ meets the line $BC$ at $M$, what is $|CM|$? $ \textbf{(A)}\ 12 \qquad\textbf{(B)}\ 9 \qquad\textbf{(C)}\ 7 \qquad\textbf{(D)}\ 5 \qquad\textbf{(E)}\ \text { None of above} $

Prove that the orthogonal projections of the vertex $ D $ of the tetrahedron $ ABCD $ onto the bisectors of the internal and external dihedral angles at the edges $ \overline{AB} $, $ \overline{BC} $ and $ \overline{CA} $ belong to one plane .

Let $ABCD$ be a parallelogram. The interior angle bisector of $\angle ADC$ intersects the line $BC$ in $E$, and the perpendicular bisector of the side $AD$ intersects the line $DE$ in $M$. Let $F= AM \cap BC$. Prove that: a) $DE=AF$; b) $AD\cdot AB = DE\cdot DM$. [i]Daniela and Marius Lobaza, Timisoara[/i]

Justify the following construction of the bisector of an angle with an inaccessible vertex: [img]https://cdn.artofproblemsolving.com/attachments/9/d/be4f7799d58a28cab3b4c515633b0e021c1502.png[/img] $M \in a$ and $N \in b$ are taken, the $4$ bisectors of the $4$ internal angles formed by $MN$ are traced with $a$ and $ b$. Said bisectors intersect at $P$ and $Q$, then $PQ$ is the bisector sought.

Given triangle $ABC$ with $|AC| > |BC|$. The point $M$ lies on the angle bisector of angle $C$, and $BM$ is perpendicular to the angle bisector. Prove that the area of triangle AMC is half of the area of triangle $ABC$. [img]https://cdn.artofproblemsolving.com/attachments/4/2/1b541b76ec4a9c052b8866acbfea9a0ce04b56.png[/img]

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.

In the triangle $ABC$, the angle - bisector $AD$ ($D \in BC$) and the median $BE$ ($E \in AC$) intersect at point $P$. Lines $AB$ and $CP$ intesect at point $F$. The parallel through $B$ to $CF$ intersects $DF$ at point $M$. Prove that $DM = BF$

Angle bisectors of angles by vertices $A$, $B$ and $C$ in triangle $ABC$ intersect opposing sides in points $A_1$, $B_1$ and $C_1$, respectively. Let $M$ be an arbitrary point on one of the lines $A_1B_1$, $B_1C_1$ and $C_1A_1$. Let $M_1$, $M_2$ and $M_3$ be orthogonal projections of point $M$ on lines $BC$, $CA$ and $AB$, respectively. Prove that one of the lines $MM_1$, $MM_2$ and $MM_3$ is equal to sum of other two

On the bisector of the angle $ BAC $ of the triangle $ ABC $ we choose the points $ {{B} _ {1}}, \, \, {{C} _ {1}} $ for which $ B {{B} _ {1 }}\perp AB $, $ C {{C} _ {1}} \perp AC $. The point $ M $ is the midpoint of the segment $ {{B} _ {1}} {{C} _ {1}} $. Prove that $ MB = MC $.