Found problems: 353
Points $A$ and $B$ are fixed on a circle $c_1$. Circle $c_2$, whose centre lies on $c_1$, touches line $AB$ at $B$. Another line through $A$ intersects $c_2$ at points $D$ and $E$, where $D$ lies between $A$ and $E$. Line $BD$ intersects $c_1$ again at $F$. Prove that line $EB$ is tangent to $c_1$ if and only if $D$ is the midpoint of the segment $BF$.
Two circles intersect at points $M$ and $N$. Let $A$ be a point on the first circle, distinct from $M,N$. The lines $AM$ and $AN$ meet the second circle again at $B$ and $C$, respectively. Prove that the tangent to the first circle at $A$ is parallel to $BC$.
Let $\omega$ be the semicircle on diameter $AB$. A line parallel to $AB$ intersects $\omega$ at $C$ and $D$ so that $B$ and $C$ lie on opposite sides of $AD$. The line through $C$ parallel to $AD$ meets $\omega$ again in $E$. Lines $BE$ and $CD$ meet in $F$ and the line through $F$ parallel to $AD$ meets $AB$ in $P$. Prove that $PC$ is tangent to $\omega$.
Suppose a triangle $\vartriangle ABC$ with $\angle C = 90^o$ is given. Let $D$ be the midpoint of $AC$, and let $E$ be the foot of the altitude through $C$ on $BD$. Show that the tangent in $C$ of the circumcircle of $\vartriangle AEC$ is perpendicular to $AB$.
Let $ABCD$ be a quadrilateral inscribed in the circle $(O)$. Let $(K)$ be an arbitrary circle passing through $B,C$. Circle $(O_1)$ tangent to $AB,AC$ and is internally tangent to $(K)$. Circle $(O_2)$ touches $DB,DC$ and is internally tangent to $(K)$. Prove that one of the two external common tangents of $(O_1)$ and $(O_2)$ is parallel to $AD$.
Trần Quang Hùng
Let $k$ be a circle with center $M.$ There is another circle $k_1$ whose center $M_1$ lies on $k,$ and we denote the line through $M$ and $M_1$ by $g.$ Let $T$ be a point on $k_1$ and inside $k.$ The tangent $t$ to $k_1$ at $T$ intersects $k$ in two points $A$ and $B.$ Denote the tangents (diifferent from $t$) to $k_1$ passing through $A$ and $B$ by $a$ and $b$, respectively. Prove that the lines $a,b,$ and $g$ are either concurrent or parallel.
Let $I$ be the incenter of a triangle $ABC$. The circle passing through $I$ and centered at $A$ meets the circumference of the triangle $ABC$ at points $M$ and $N$. Prove that the line $MN$ touches the incircle of the triangle $ABC$.
(I. Kachan)
Given a circle and a point $P$ inside it, different from the center. We consider pairs of circles tangent to the given internally and to each other at point $P$. Find the locus of the points of intersection of the common external tangents to these circles.
Let $ABC$ be an acute triangle with $AB < AC$, and let $BE$ and $CF$ be the altitudes. Let the median $AM$ intersect $BE$ at point $P$, and let line $CP$ intersect $AB$ at point $D$ (see Figure 2). Prove that $DE \parallel BC$, and $AC$ is tangent to the circumcircle of $\vartriangle DEF$.
[img]https://cdn.artofproblemsolving.com/attachments/f/7/bbad9f6019a77c6aa46c3a821857f06233cb93.png[/img]
$AD$ and $BE$ are the altitudes of the triangle $ABC$. It turned out that the point $C'$, symmetric to the vertex $C$ wrt to the midpoint of the segment $DE$, lies on the side $AB$. Prove that $AB$ is tangent to the circle circumscribed around the triangle $DEC'$.
Let the two tangents from a point $A$ outside a circle $k$ touch $k$ at $M$ and $N$. A line $p$ through $A$ intersects $k$ at $B$ and $C$, and $D$ is the midpoint of $MN$. Prove that $MN$ bisects the angle $BDC$
A line through point $P$ outside of circle $O$ meets the said circle at $B,C$ ($PB < PC$). Let $PO$ meet circle $O$ at $Q,D$ (with $PQ < PD$). Let the line passing $Q$ and perpendicular to $BC$ meet circle $O$ at $A$. If $BD^2 = AD\cdot CP$, prove that $PA$ is a tangent to $O$.
The center of the inscribed circle of triangle $ABC$ is $I$. Let $e$ be the perpendicular line on $CI$ passing through $I$. The line $e$ itnersects the side $AC$ at $A'$ and the side $BC$ at point $B'$. Let $A''$ be the symmetric point of $A$ wrt $A'$, $B''$ be the symmetric point of $B$ wrt $B'$. Prove that $A''B''$ is a line tangent to the incircle.
Let $k$ be a circle centered at $O$. Let $\overline{AB}$ be a chord of that circle and $M$ its midpoint. Tangent on $k$ at points $A$ and $B$ intersect at $T$. The line $\ell$ goes through $T$, intersect the shorter arc $AB$ at the point $C$ and the longer arc $AB$ at the point $D$, so that $|BC| = |BM|$. Prove that the circumcenter of the triangle $ADM$ is the reflection of $O$ across the line $AD$
Two non-coplanar circles in space are tangent at a point and have the same tangents at this point. Show that both circles lie on some sphere.
The incircle of triangle $ABC$ tangents $BC$, $CA$, $AB$ at $D$, $E$, $F$, respectively. Let the tangents of $E$, $F$ with respect to $\odot(AEF)$ intersect at $P$, and $X$ be a point on $BC$ such that $EF$, $DP$, $AX$ are concurrent. Define $Q$, $Y$ and $R$, $Z$ similarly. Show that $X$, $Y$, $Z$ are collinear.
In an acute-angled triangle $ABC$, $M$ is a point on the side $BC$, the line $AM$ meets the circumcircle $\omega$ of $ABC$ at the point $Q$ distinct from $A$. The tangent to $\omega$ at $Q$ intersects the line through $M$ perpendicular to the diameter $AK$ of $\omega$ at the point $P$. Let $L$ be the point on $\omega$ distinct from $Q$ such that $PL$ is tangent to $\omega$ at $L$. Prove that $L,M$ and $K$ are collinear.
Let $ABC$ be an obtuse triangle with $\angle A > \angle B > \angle C$, and let $M$ be a midpoint of the side $BC$. Let $D$ be a point on the arc $AB$ of the circumcircle of triangle $ABC$ not containing $C$. Suppose that the circle tangent to $BD$ at $D$ and passing through $A$ meets the circumcircle of triangle $ABM$ again at $E$ and $\overline{BD}=\overline{BE}$. $\omega$, the circumcircle of triangle $ADE$, meets $EM$ again at $F$.
Prove that lines $BD$ and $AE$ meet on the line tangent to $\omega$ at $F$.
Let $ABC$ be an isosceles triangle with $AC = BC$ and circumcircle $k$. The point $D$ lies on the shorter arc of $k$ over the chord $BC$ and is different from $B$ and $C$. Let $E$ denote the intersection of $CD$ and $AB$. Prove that the line through $B$ and $C$ is a tangent of the circumcircle of the triangle $BDE$.
(Karl Czakler)
Let $ABC$ be a triangle with $AB = AC$. With center in a point of the side $BC$, the circle $S$ is constructed that is tangent to the sides $AB$ and $AC$. Let $P$ and $Q$ be any points on the sides $AB$ and $AC$ respectively, such that $PQ$ is tangent to $S$. Show that $PB \cdot CQ = \left(\frac{BC}{2}\right)^2$
$C$ is a point on the semicircle diameter $AB$, between $A$ and $B$. $D$ is the foot of the perpendicular from $C$ to $AB$. The circle $K_1$ is the incircle of $ABC$, the circle $K_2$ touches $CD,DA$ and the semicircle, the circle $K_3$ touches $CD,DB$ and the semicircle. Prove that $K_1,K_2$ and $K_3$ have another common tangent apart from $AB$.
Let $ABC$ be a right triangle with hypotenuse $BC$. The tangent to the circumcircle of triangle $ABC$ at $A$ intersects the line $BC$ at $T$. The points $D$ and $E$ are chosen so that $AD = BD, AE = CE,$ and $\angle CBD = \angle BCE < 90^{\circ}$. Prove that $D, E,$ and $T$ are collinear.
Proposed by [i]Nikola Velov, Macedonia[/i]
Let $AM$ and $AN$ be the tangents to a circle $\Gamma$ drawn from a point $A$ ($M$ and $N$ lie on the circle). A line passing through $A$ cuts $\Gamma$ at $B$ and $C$, with B between $A$ and $C$ such that $AB: BC = 2: 3$. If $P$ is the intersection point of $AB$ and $MN$, calculate the ratio $AP: CP$ .
Let $ABC$ be a triangle with $\angle ABC \neq 90^\circ$ and $AB$ its shortest side. Denote by $H$ the intersection of the altitudes of triangle $ABC$. Let $K$ be the circle through $A$ with centre $B$. Let $D$ be the other intersection of $K$ and $AC$. Let $K$ intersect the circumcircle of $BCD$ again at $E$. If $F$ is the intersection of $DE$ and $BH$, show that $BD$ is tangent to the circle through $D$, $F$, and $H$.
Convex quadrilateral $ABCD$ has $\angle ABC = \angle CDA = 90^{\circ}$. Point $H$ is the foot of the perpendicular from $A$ to $BD$. Points $S$ and $T$ lie on sides $AB$ and $AD$, respectively, such that $H$ lies inside triangle $SCT$ and \[
\angle CHS - \angle CSB = 90^{\circ}, \quad \angle THC - \angle DTC = 90^{\circ}. \] Prove that line $BD$ is tangent to the circumcircle of triangle $TSH$.