Found problems: 63
The two circles ${{w} _ {1}}, \, \, {{w} _ {2}}$ touch externally at the point $Q$. The common external tangent of these circles is tangent to ${{w} _ {1}}$ at the point $B$, $BA$ is the diameter of this circle. A tangent to the circle ${{w} _ {2}} $ is drawn through the point $A$, which touches this circle at the point $C$, such that the points $B$ and $C$ lie in one half-plane relative to the line $AQ$. Prove that the circle ${{w} _ {1}}$ bisects the segment $C $.
(Igor Nagel)
A circle touches sides $AB, BC, CD$ of a parallelogram $ABCD$ at points $K, L, M$ respectively. Prove that the line $KL$ bisects the height of the parallelogram drawn from the vertex $C$ to $AB$.
Let $ABC$ be a triangle with $M, N, P$ as midpoints of the segments $BC, CA,AB$ respectively. Suppose that $I$ is the intersection of angle bisectors of $\angle BPM, \angle MNP$ and $J$ is the intersection of angle bisectors of $\angle CN M, \angle MPN$. Denote $(\omega_1)$ as the circle of center $I$ and tangent to $MP$ at $D$, $(\omega_2)$ as the circle of center $J$ and tangent to $MN$ at $E$.
a) Prove that $DE$ is parallel to $BC$.
b) Prove that the radical axis of two circles $(\omega_1), (\omega_2)$ bisects the segment $DE$.
The altitudes $AA_1, BB_1$ and $CC_1$ were drawn in the triangle $ABC$. Point $K$ is a projection of point $B$ on $A_1C_1$. Prove that the symmmedian $\vartriangle ABC$ from the vertex $B$ divides the segment $B_1K$ in half.
(Anton Trygub)
The altitudes $AN$ and $BM$ are drawn in triangle $ABC$. Prove that the perpendicular bisector to the segment $NM$ divides the segment $AB$ in half.
Given acute triangle $\vartriangle ABC$ with orthocenter $H$ and circumcenter $O$ ($O \ne H$) . Let $\Gamma$ be the circumcircle of $\vartriangle BOC$ . Segment $OH$ untersects $\Gamma$ at point $P$. Extension of $AO$ intersects $\Gamma$ at point $K$. If $AP \perp OH$, prove that $PK$ bisects $BC$.
[img]https://cdn.artofproblemsolving.com/attachments/a/b/267053569c41692f47d8f4faf2a31ebb4f4efd.png[/img]
Let $\odot O$ be the inscribed circle of $\vartriangle ABC$, with $D$, $E$, $N$ the touchpoints with sides $AB$, $AC$, $BC$ respectively. Extension of $NO$ intersects segment $DE$ at point $K$. Extension of $AK$ intersects segment $BC$ at point $M$. Prove that $M$ is the midpoint of $BC$.
[img]https://cdn.artofproblemsolving.com/attachments/a/6/a503c500178551ddf9bdb1df0805ed22bc417d.png[/img]
Given a triangle $ABC$, the perpendicular bisector of the side $AC$ intersects the angle bisector of the triangle $AK$ at the point $P$, $M$ - such a point that $\angle MAC = \angle PCB$, $\angle MPA = \angle CPK$, and points $M$ and $K$ lie on opposite sides of the line $AC$. Prove that the line $AK$ bisects the segment $BM$.
(Anton Trygub)
The two circles ${{w} _ {1}}, \, \, {{w} _ {2}}$ touch externally at the point $Q$. The common external tangent of these circles is tangent to ${{w} _ {1}}$ at the point $B$, $BA$ is the diameter of this circle. A tangent to the circle ${{w} _ {2}} $ is drawn through the point $A$, which touches this circle at the point $C$, such that the points $B$ and $C$ lie in one half-plane relative to the line $AQ$. Prove that the circle ${{w} _ {1}}$ bisects the segment $C $.
(Igor Nagel)
Let $O$ be the center of the circle of the triangle $ABC$ with $AB \ne AC$. Furthermore, let $S$ and $T$ be points on the rays $AB$ and $AC$, such that $\angle ASO = \angle ACO$ and $\angle ATO = \angle ABO$. Show that $ST$ bisects the segment $BC$.
Let $ABC$ be a triangle with $AB \ne AC$ and incenter $I$. The incircle touches $BC$ at $D$. Let $M$ be the midpoint of $BC$ . Show that the line $IM$ bisects segment $AD$ .
Let $ABC$ be an acute, non-isosceles triangle with circumcenter $O$, incenter $I$ and $(I)$ tangent to $BC$, $CA$, $AB$ at $D, E, F$ respectively. Suppose that $EF$ cuts $(O)$ at $P, Q$. Prove that $(PQD)$ bisects segment $BC$.
Let $ABC$ be a triangle with $|AB|> |BC|$. Let $D$ be the midpoint of $AC$. Let $E$ be the intersection of the angular bisector of $\angle ABC$ and the line $AC$. Let $F$ be the point on $BE$ such that $CF$ is perpendicular to $BE$. Finally, let $G$ be the intersection of $CF$ and $BD$. Prove that $DF$ divides the line segment $EG$ into two equal parts.