A semicircle $(O)$ is drawn with the center $O$, where $O$ lies on a line $\ell$. $C$ and $D$ lie on the circle $(O)$, and the tangent lines of $(O)$ at points $C$ and $D$ intersects the line $\ell$ at points $B$ and $A$, respectively, such that $O$ lies between points $B$ and $A$. Let $E$ be the intersection point between $AC$ and $BD$, and $F$ the point on $\ell$ so that $EF $ is perpendicular to line $\ell$. Prove that $EF$ bisects the angle $\angle CFD$.

Given $\triangle ABC$ with $AB < AC$, let $\omega$ be the circle centered at the midpoint $M$ of $BC$ with diameter $AC - AB$. The internal bisector of $\angle BAC$ intersects $\omega$ at distinct points $X$ and $Y$. Let $T$ be the point on the plane such that $TX$ and $TY$ are tangent to $\omega$. Prove that $AT$ is perpendicular to $BC$.

Let $ABCD$ be a convex quadrileteral such that $m(\widehat{ABD})=40^\circ$, $m(\widehat{DBC})=70^\circ$, $m(\widehat{BDA})=80^\circ$, and $m(\widehat{BDC})=50^\circ$. What is $m(\widehat{CAD})$? $ \textbf{(A)}\ 25^\circ \qquad\textbf{(B)}\ 30^\circ \qquad\textbf{(C)}\ 35^\circ \qquad\textbf{(D)}\ 38^\circ \qquad\textbf{(E)}\ 40^\circ $

In triangle $ABC$, the midpoints of sides $AB$ and $AC$ are $D$ and $E$, respectively. Prove that the bisectors of the angles $BDE$ and $CED$ intersect at the side $BC$ if the length of side $BC$ is the arithmetic mean of the lengths of sides $AB$ and $AC$.

The feet of the perpendiculars from the intersection point of the diagonals of a convex cyclic quadrilateral to the sides form a quadrilateral $q$. Show that the sum of the lengths of each pair of opposite sides of $q$ is equal.

Let $\omega,\omega_1,\omega_2$ be three mutually tangent circles such that $\omega_1,\omega_2$ are externally tangent at $P$, $\omega_1,\omega$ are internally tangent at $A$, and $\omega,\omega_2$ are internally tangent at $B$. Let $O,O_1,O_2$ be the centers of $\omega,\omega_1,\omega_2$, respectively. Given that $X$ is the foot of the perpendicular from $P$ to $AB$, prove that $\angle{O_1XP}=\angle{O_2XP}$. [i]David Yang.[/i]

Let $ABC$ be an isosceles triangle with $AB = AC$, $P$ be the midpoint of the minor arc $AB$ of its circumcircle, and $Q$ be the midpoint of $AC$. A circumcircle of triangle $APQ$ centered at $O$ meets $AB$ for the second time at point $K$. Prove that lines $PO$ and $KQ$ meet on the bisector of angle $ABC$.

In a triangle $ABC$, let $AA', BB'$ and $CC'$ be the bisectors. Suppose $A'B' \cap CC' =P$ and $A'C' \cap BB'= Q$. Prove that $\angle PAC = \angle QAB$.

Let $ABCD$ be a cyclic quadrilateral. Let $P$, $Q$, $R$ be the feet of the perpendiculars from $D$ to the lines $BC$, $CA$, $AB$, respectively. Show that $PQ=QR$ if and only if the bisectors of $\angle ABC$ and $\angle ADC$ are concurrent with $AC$.

The angle $B$ of a triangle $ABC$ is $120^o$. Let $M$ be a point on the side $AC$ and $K$ a point on the extension of the side $AB$, such that $BM$ is the internal bisector of the angle $\angle ABC$ and $CK$ is the external bisector corresponding to the angle $\angle ACB$ . The segment $MK$ intersects $BC$ at point $P$. Prove that $\angle APM = 30^o$.

Let $\Gamma_1$ and $\Gamma_2$ be circles internally tangent at point $A$, with $\Gamma_1$ inside $\Gamma_2$. Let $BC$ be a chord of $\Gamma_2$ which is tangent to $\Gamma_1$ at point $D$. Prove that line $AD$ is the angle bisector of $\angle BAC$.

In $\triangle ABC$ shown in the figure, $AB=7$, $BC=8$, $CA=9$, and $\overline{AH}$ is an altitude. Points $D$ and $E$ lie on sides $\overline{AC}$ and $\overline{AB}$, respectively, so that $\overline{BD}$ and $\overline{CE}$ are angle bisectors, intersecting $\overline{AH}$ at $Q$ and $P$, respectively. What is $PQ$? [asy]draw((0,0)--(7,0)); draw((0,0)--(33/7,7.66651)); draw((33/7,7.66651)--(7,0)); draw((11/5,7*7.66651/15)--(7,0)); draw((63/17,0)--(33/7,7.66651)); draw((0,0)--(45/7,7.66651/4)); dot((0,0)); label("A",(0,0),SW); dot((7,0)); label("B",(7,0),SE); dot((33/7,7.66651)); label("C",(33/7,7.66651),N); dot((11/5,7*7.66651/15)); label("D",(11/5+.2,7*7.66651/15-.25),S); dot((63/17,0)); label("E",(63/17,0),NE); dot((45/7,7.66651/4)); label("H",(44/7,7.66651/4),NW); dot((27/7,3*7.66651/20)); label("P",(27/7,3*7.66651/20),NW); dot((5,7*7.66651/36)); label("Q",(5,7*7.66651/36),N); label("9",(33/14,7.66651/2),NW); label("8",(41/7,7.66651/2),NE); label("7",(3.5,0),S);[/asy] $\textbf{(A)}\ 1 \qquad \textbf{(B)}\ \frac{5}{8}\sqrt{3} \qquad \textbf{(C)}\ \frac{4}{5}\sqrt{2} \qquad \textbf{(D)}\ \frac{8}{15}\sqrt{5} \qquad \textbf{(E)}\ \frac{6}{5}$

Let $ABCD$ be a convex quadrilateral and let $P$ and $Q$ be the points on the sides $AD$ and $BC$ respectively such that $\frac{AP}{PD}=\frac{BQ}{QC}=\frac{AB}{CD}$. Prove that the line $PQ$ forms equal angles with the lines $AB$ and $CD$.

We consider the quadrilateral $ABCD$, with $AD = CD = CB$ and $AB \parallel CD$. Points $E$ and $F$ belong to the segments $CD$ and $CB$ so that angles $\angle ADE = \angle AEF$. Prove that: a) $4CF \le CB$ , b) if $4CF = CB$, then $AE$ is the bisector of the angle $\angle DAF$.

In a triangle $ \vartriangle ABC $, let $ h_a, h_b $ and $ h_c $ the atlitudes. Let $ D $ be the point where the inner bisector of $ \angle BAC $ cuts to the side $ BC $ and $ d_a $ is the distance from the $ D $ point next to $ AB $. The distances $ d_b $ and $ d_c $ are similarly defined. Show that: $$ \dfrac {3} {2} \le \dfrac {d_a} {h_a} + \dfrac {d_b} {h_b} + \dfrac {d_c} {h_c} $$ For what kind of triangles does the equality hold?

A circle with center O is inscribed in a quadrilateral ABCD and touches its non-parallel sides BC and AD at E and F respectively. The lines AO and DO meet the segment EF at K and N respectively, and the lines BK and CN meet at M. Prove that the points O,K,M and N lie on a circle.

In an acute triangle $ABC$ the angle $C$ is greater than the angle $A$. Let $AE$ be a diameter of the circumcircle of the triangle. Let the intersection point of the ray $AC$ and the tangent of the circumcircle through the vertex $B$ be $K$. The perpendicular to $AE$ through $K$ intersects the circumcircle of the triangle $BCK$ for the second time at point $D$. Prove that $CE$ bisects the angle $BCD$.

By straightedge and compass, reconstruct a right triangle $ABC$ ($\angle C = 90^o$), given the vertices $A, C$ and a point on the bisector of angle $B$.

$ C$ and $ D$ are points on a semicircle. The tangent at $ C$ meets the extended diameter of the semicircle at $ B$, and the tangent at $ D$ meets it at $ A$, so that $ A$ and $ B$ are on opposite sides of the center. The lines $ AC$ and $ BD$ meet at $ E$. $ F$ is the foot of the perpendicular from $ E$ to $ AB$. Show that $ EF$ bisects angle $ CFD$

Two equal non-intersecting wooden disks, one gray and one black, are glued to a plane. A triangle with one gray side and one black side can be moved along the plane so that the disks remain outside the triangle, while the colored sides of the triangle are tangent to the disks of the same color (the tangency points are not the vertices). Prove that the line that contains the bisector of the angle between the gray and black sides always passes through some fixed point of the plane. (Egor Bakaev, Pavel Kozhevnikov, Vladimir Rastorguev) (Senior version[url=https://artofproblemsolving.com/community/c6h2102856p15209040] here[/url])

In the quadrilateral $ABCD$ the condition $AD = AB + CD$ is fulfilled. The bisectors of the angles $BAD$ and $ADC$ intersect at the point $P $, as shown in Fig. Prove that $BP = CP$. [img]https://cdn.artofproblemsolving.com/attachments/3/1/67268635aaef9c6dc3363b00453b327cbc01f3.png[/img] (Maria Rozhkova)

In $\triangle {ABC}$, ${D}$ is on the internal angle bisector of $\angle BAC$ and $\angle ADB=\angle ACD$. $E, F$ is on the external angle bisector of $\angle BAC$, such that $AE=BE$ and $AF=CF$. The circumcircles of $\triangle ACE$ and $\triangle ABF$ intersects at ${A}$ and ${K}$ and $A'$ is the reflection of ${A}$ with respect to $BC$. Prove that: if $AD=BC$, then the circumcenter of $\triangle AKA'$ is on line $AD$.

In the acute triangle $ABC$, the midpoint of side $BC$ is called $M$. Point $X$ lies on the angle bisector of $\angle AMB$ such that $\angle BXM = 90^o$. Point $Y$ lies on the angle bisector of $\angle AMC$ such that $\angle CYM = 90^o$. Line segments $AM$ and $XY$ intersect in point $Z$. Prove that $Z$ is the midpoint of $XY$ . [asy] import geometry; unitsize (1.2 cm); pair A, B, C, M, X, Y, Z; A = (0,0); B = (4,1.5); C = (0.5,3); M = (B + C)/2; X = extension(M, incenter(A,B,M), B, B + rotate(90)*(incenter(A,B,M) - M)); Y = extension(M, incenter(A,C,M), C, C + rotate(90)*(incenter(A,C,M) - M)); Z = extension(A,M,X,Y); draw(A--B--C--cycle); draw(A--M); draw(M--interp(M,X,2)); draw(M--interp(M,Y,2)); draw(B--X, dotted); draw(C--Y, dotted); draw(X--Y); dot("$A$", A, SW); dot("$B$", B, E); dot("$C$", C, N); dot("$M$", M, NE); dot("$X$", X, NW); dot("$Y$", Y, NE); dot("$Z$", Z, S); [/asy]

Let ABCD be a parallelogram, M the midpoint of AB and N the intersection of CD and the angle bisector of ABC. Prove that CM and BN are perpendicular iff AN is the angle bisector of DAB.

In an acute triangle $\triangle{ABC}$, let $AE$ and $BF$ be highs of it, and $H$ its orthocenter. The symmetric line of $AE$ with respect to the angle bisector of $\sphericalangle{A}$ and the symmetric line of $BF$ with respect to the angle bisector of $\sphericalangle{B}$ intersect each other on the point $O$. The lines $AE$ and $AO$ intersect again the circuncircle to $\triangle{ABC}$ on the points $M$ and $N$ respectively. Let $P$ be the intersection of $BC$ with $HN$; $R$ the intersection of $BC$ with $OM$; and $S$ the intersection of $HR$ with $OP$. Show that $AHSO$ is a paralelogram.