Let $ABC$ be a triangle with $\angle A=60^o, AB\ne AC$ and let $AD$ be the angle bisector of $\angle A$. Line $(e)$ that is perpendicular on the angle bisector $AD$ at point $A$, intersects the extension of side $BC$ at point $E$ and also $BE=AB+AC$. Find the angles $\angle B$ and $\angle C$ of the triangle $ABC$.

In triangle $ABC$, we have $\angle ABC=60$. The line through $B$ perpendicular to side $AB$ intersects angle bisector of $\angle BAC$ in $D$ and the line through $C$ perpendicular $BC$ intersects angle bisector of $\angle ABC$ in $E$. prove that $\angle BED\le 30$.

$AD$ is the altitude on side $BC$ of triangle $ABC$. If $BC+AD-AB-AC = 0$, find the range of $\angle BAC$. [i]Alternative formulation.[/i] Let $AD$ be the altitude of triangle $ABC$ to the side $BC$. If $BC+AD=AB+AC$, then find the range of $\angle{A}$.

Let $D$, $E$, $F$ be the midpoints of the arcs $BC$, $CA$, $AB$ on the circumcircle of a triangle $ABC$ not containing the points $A$, $B$, $C$, respectively. Let the line $DE$ meets $BC$ and $CA$ at $G$ and $H$, and let $M$ be the midpoint of the segment $GH$. Let the line $FD$ meet $BC$ and $AB$ at $K$ and $J$, and let $N$ be the midpoint of the segment $KJ$. a) Find the angles of triangle $DMN$; b) Prove that if $P$ is the point of intersection of the lines $AD$ and $EF$, then the circumcenter of triangle $DMN$ lies on the circumcircle of triangle $PMN$.

Let $ ABCD$ be a quadrilateral which is a cyclic quadrilateral and a tangent quadrilateral simultaneously. (By a [i]tangent quadrilateral[/i], we mean a quadrilateral that has an incircle.) Let the incircle of the quadrilateral $ ABCD$ touch its sides $ AB$, $ BC$, $ CD$, and $ DA$ in the points $ K$, $ L$, $ M$, and $ N$, respectively. The exterior angle bisectors of the angles $ DAB$ and $ ABC$ intersect each other at a point $ K^{\prime}$. The exterior angle bisectors of the angles $ ABC$ and $ BCD$ intersect each other at a point $ L^{\prime}$. The exterior angle bisectors of the angles $ BCD$ and $ CDA$ intersect each other at a point $ M^{\prime}$. The exterior angle bisectors of the angles $ CDA$ and $ DAB$ intersect each other at a point $ N^{\prime}$. Prove that the straight lines $ KK^{\prime}$, $ LL^{\prime}$, $ MM^{\prime}$, and $ NN^{\prime}$ are concurrent.

Let $ABC$ be a triangle with circumcentre $O$. The points $D,E,F$ lie in the interiors of the sides $BC,CA,AB$ respectively, such that $DE$ is perpendicular to $CO$ and $DF$ is perpendicular to $BO$. (By interior we mean, for example, that the point $D$ lies on the line $BC$ and $D$ is between $B$ and $C$ on that line.) Let $K$ be the circumcentre of triangle $AFE$. Prove that the lines $DK$ and $BC$ are perpendicular. [i]Netherlands (Merlijn Staps)[/i]

Given an acute triangle $ ABC$. The incircle of triangle $ ABC$ touches $ BC,CA,AB$ respectively at $ D,E,F$. The angle bisector of $ \angle A$ cuts $ DE$ and $ DF$ respectively at $ K$ and $ L$. Suppose $ AA_1$ is one of the altitudes of triangle $ ABC$, and $ M$ be the midpoint of $ BC$. (a) Prove that $ BK$ and $ CL$ are perpendicular with the angle bisector of $ \angle BAC$. (b) Show that $ A_1KML$ is a cyclic quadrilateral.

Let $AC>BC$ in $\triangle ABC$ and $M$ and $N$ be the midpoints of $AC$ and $BC$ respectively. The angle bisector of $\angle B$ intersects $\overline{MN}$ at $P$. The incircle of $\triangle ABC$ has center $I$ and touches $BC$ at $Q$. The perpendiculars from $P$ and $Q$ to $MN$ and $BC$ respectively intersect at $R$. Let $S=AB\cap RN$. a) Prove that $PCQI$ is cyclic b) Express the length of the segment $BS$ with $a$, $b$, $c$ - the side lengths of $\triangle ABC$ .

Let $ABC$ be a triangle, $P$ and $Q$ the intersections of the parallel line to $BC$ that passes through $A$ with the external angle bisectors of angles $B$ and $C$, respectively. The perpendicular to $BP$ at $P$ and the perpendicular to $CQ$ at $Q$ meet at $R$. Let $I$ be the incenter of $ABC$. Show that $AI = AR$.

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]

Let $ABC$ be a triangle with $AB < AC$ and incentre $I$. Let $E$ be the point on the side $AC$ such that $AE = AB$. Let $G$ be the point on the line $EI$ such that $\angle IBG = \angle CBA$ and such that $E$ and $G$ lie on opposite sides of $I$. Prove that the line $AI$, the line perpendicular to $AE$ at $E$, and the bisector of the angle $\angle BGI$ are concurrent.

In $\triangle ABC$, $M$ is the midpoint of side $BC$, $AN$ bisects $\angle BAC$, $BN\perp AN$ and $\theta$ is the measure of $\angle BAC$. If sides $AB$ and $AC$ have lengths $14$ and $19$, respectively, then length $MN$ equals [asy] size(230); defaultpen(linewidth(0.7)+fontsize(10)); pair B=origin, A=14*dir(36), C=intersectionpoint(B--(9001,0), Circle(A,19)), M=midpoint(B--C), b=A+14*dir(A--C), N=foot(A, B, b); draw(N--B--A--N--M--C--A^^B--M); markscalefactor=0.1; draw(rightanglemark(B,N,A)); pair point=N; label("$A$", A, dir(point--A)); label("$B$", B, dir(point--B)); label("$C$", C, dir(point--C)); label("$M$", M, S); label("$N$", N, dir(30)); label("$19$", (A+C)/2, dir(A--C)*dir(90)); label("$14$", (A+B)/2, dir(A--B)*dir(270)); [/asy] $\displaystyle \text{(A)} \ 2 \qquad \text{(B)} \ \frac{5}{2} \qquad \text{(C)} \ \frac{5}{2} - \sin \theta \qquad \text{(D)} \ \frac{5}{2} - \frac{1}{2} \sin \theta \qquad \text{(E)} \ \frac{5}{2} - \frac{1}{2} \sin \left(\frac{1}{2} \theta\right)$

In the triangle $ABC$ we have $|AB|<|AC|$. The bisectors of the angles at $B$ and $C$ meet $AC$ and $AB$ at $D$ and $E$ respectively. $BD$ and $CE$ intersect at the incenter $I$ of $\triangle ABC$. Prove that $\angle BAC=60^\circ$ if and only if $|IE|=|ID|$

Let $ABC$ be a triangle such that it's circumcircle radius is equal to the radius of outer inscribed circle with respect to $A$. Suppose that the outer inscribed circle with respect to $A$ touches $BC,AC,AB$ at $M,N,L$. Prove that $O$ (Center of circumcircle) is the orthocenter of $MNL$.

In triangle $ABC$, $AB = 52$, $BC = 56$, $CA = 60$. Let $D$ be the foot of the altitude from $A$ and $E$ be the intersection of the internal angle bisector of $\angle BAC$ with $BC$. Find $DE$.

Let $ABC$ be a right triangle with $m(\widehat{A})=90^\circ$. Let $APQR$ be a square with area $9$ such that $P\in [AC]$, $Q\in [BC]$, $R\in [AB]$. Let $KLMN$ be a square with area $8$ such that $N,K\in [BC]$, $M\in [AB]$, and $L\in [AC]$. What is $|AB|+|AC|$? $ \textbf{(A)}\ 8 \qquad\textbf{(B)}\ 10 \qquad\textbf{(C)}\ 12 \qquad\textbf{(D)}\ 14 \qquad\textbf{(E)}\ 16 $

Given triangle $ABC$, $AL$ bisects angle $\angle BAC$ with $L$ on side $BC$. Lines $LR$ and $LS$ are parallel to $BA$ and $CA$ respectively, $R$ on side $AC$ and$ S$ on side $AB$, respectively. Through point $B$ draw a perpendicular on $AL$, intersecting $LR$ at $M$. If point $D$ is the midpoint of $BC$, prove that that the three points $A, M, D$ lie on a straight line.

Let $ ABC $ be a triangle and the point $ D $ is on the segment $ BC $ such that $ AD $ is the interior bisector of $ \angle A $. We stretch $ AD $ such that it meets the circumcircle of $ \Delta ABC $ at $ M $. We draw a line from $ D $ such that it meets the lines $ MB,MC $ at $ P,Q $, respectively ($ M $ is not between $ B,P $ and also is not between $ C,Q $). Prove that $ \angle PAQ\geq\angle BAC $.

$ABC$ is a right-angled triangle, with $\angle ABC = 90^{\circ}$. $B'$ is the reflection of $B$ over $AC$. $M$ is the midpoint of $AC$. We choose $D$ on $\overrightarrow{BM}$, such that $BD = AC$. Prove that $B'C$ is the angle bisector of $\angle MB'D$. NOTE: An important condition not mentioned in the original problem is $AB<BC$. Otherwise, $\angle MB'D$ is not defined or $B'C$ is the external bisector.

Circle $k$ and its diameter $AB$ are given. Find the locus of the centers of circles inscribed in the triangles having one vertex on $AB$ and two other vertices on $k.$

Let $ABC$ be an acute non isosceles triangle. The angle bisector of angle $A$ meets again the circumcircle of the triangle $ABC$ in $D$. Let $O$ be the circumcenter of the triangle $ABC$. The angle bisectors of $\angle AOB$, and $\angle AOC$ meet the circle $\gamma$ of diameter $AD$ in $P$ and $Q$ respectively. The line $PQ$ meets the perpendicular bisector of $AD$ in $R$. Prove that $AR // BC$.

Let $\Delta ABC$ be a triangle with an inscribed circle centered at $I$. The line perpendicular to $AI$ at $I$ intersects $\odot (ABC)$ at $P,Q$ such that, $P$ lies closer to $B$ than $C$. Let $\odot (BIP) \cap \odot (CIQ) =S$. Prove that, $SI$ is the angle bisector of $\angle PSQ$

In a triangle $ABC$, the median $AD$ (with $D$ on $BC$) and the angle bisector $BE$ (with $E$ on $AC$) are perpedicular to each other. If $AD = 7$ and $BE = 9$, find the integer nearest to the area of triangle $ABC$.

Let $ABCD$ be a rectangle and denote by $M$ and $N$ the midpoints of $AD$ and $BC$ respectively. The point $P$ is on $(CD$ such that $D\in (CP)$, and $PM$ intersects $AC$ in $Q$. Prove that $m(\angle{MNQ})=m(\angle{MNP})$.

The incircle of the triangle $ABC$ touches the sides $AC$ and $BC$ at points $P$ and $Q$ respectively. $N$ and $M$ are the midpoints of $AC$ and $BC$ respectively. Let $X=AM\cap BP, Y=BN\cap AQ$. Given $C,X,Y$ are collinear, prove that $CX$ is the angle bisector of the angle $ACB$. I. Gorodnin