Found problems: 509
In a square $ABCD$, $K$ is a point on the side $BC$ and the bisector of $\angle KAD$ cuts the side $CD$ at the point $M$. Prove that the length of segment $AK$ is equal to the sum of the lengths of segments $DM$ and $BK$.
(Folklore)
An acute-angled triangle $ABC$ is inscribed in a circle with center $O$. The bisector of $\angle A$ meets $BC$ at $D$, and the perpendicular to $AO$ through $D$ meets the segment $AC$ in a point $P$. Show that $AB = AP$.
Given a triangle $ABC$. Point $M$ moves along the side $BA$ and point $N$ moves along the side $AC$ beyond point $C$ such that $BM=CN$. Find the geometric locus of the centers of the circles circumscribed around the triangle $AMN$.
Let $ABC$ be an acute angled triangle with circumcircle $O$. Line $AO$ intersects the circumcircle of triangle $ABC$ again at point $D$. Let $P$ be a point on the side $BC$. Line passing through $P$ perpendicular to $AP$ intersects lines $DB$ and $DC$ at $E$ and $F$ respectively . Line passing through $D$ perpendicular to $BC$ intersects $EF$ at point $Q$. Prove that $EQ = FQ$ if and only if $BP = CP$.
Point $P$ is chosen inside triangle $ABC$ so that $\angle APC+\angle ABC=180^o$ and $BC=AP.$ On the side $AB$, a point $K$ is chosen such that $AK = KB + PC$. Prove that $CK \perp AB$.
In the $9$-gon $ABCDEFGHI$, all sides have equal lengths and all angles are equal. Prove that $|AB| + |AC| = |AE|$.
[img]https://cdn.artofproblemsolving.com/attachments/6/2/8c82e8a87bf8a557baaf6ac72b3d18d2ba3965.png[/img]
Let $ABC$ be a triangle with $\angle CAB = 2 \angle ABC$. Assume that a point $D$ is inside the triangle $ABC$ exists such that $AD = BD$ and $CD = AC$. Show that $\angle ACB = 3 \angle DCB$.
In an acute triangle the feet of altitudes drawn from vertices $A$ and $B$ are $D$ and $E$, respectively. Let $M$ be the midpoint of side $AB$. Line $CM$ intersects the circumcircle of $CDE$ again in point $P$ and the circumcircle of $CAB$ again in point $Q$. Prove that $|MP| = |MQ|$.
A semicircle is erected over the segment $AB$ with center $M$. Let $P$ be one point different from $A$ and $B$ on the semicircle and $Q$ the midpoint of the arc of the circle $AP$. The point of intersection of the straight line $BP$ with the parallel to $P Q$ through $M$ is $S$. Prove that $PM = PS$ holds.
[i](Karl Czakler)[/i]
Two equal circles $S_1$ and $S_2$ meet at two different points. The line $\ell$ intersects $S_1$ at points $A,C$ and $S_2$ at points $B,D$ respectively (the order on $\ell$: $A,B,C,D$) . Define circles $\Gamma_1$ and $\Gamma_2$ as follows: both $\Gamma_1$ and $\Gamma_2$ touch $S_1$ internally and $S_2$ externally, both $\Gamma_1$ and $\Gamma_2$ line $\ell$, $\Gamma_1$ and $\Gamma_2$ lie in the different halfplanes relatively to line $\ell$. Suppose that $\Gamma_1$ and $\Gamma_2$ touch each other. Prove that $AB=CD$.
I. Voronovich
Angles are equal in a hexagon, three main diagonals are equal and the other six diagonals are also equal. Is it true that it has equal sides?
Let $ABCD$ be a cyclic quadrilateral such that $|AD| =|BD|$ and let $M$ be the intersection of its diagonals. Furthermore, let $N$ be the second intersection of the diagonal $AC$ with the circle passing through points $B, M$ and the center of the circle inscribed in triangle $BCM$. Prove that $AN \cdot NC = CD \cdot BN$
In Fig. , $PA$ and $QB$ are tangents to the circle at $A$ and $B$ respectively. The line $AB$ is extended to meet $PQ$ at $S$. Suppose that $PA = QB$. Prove that $QS = SP$.
[img]https://cdn.artofproblemsolving.com/attachments/6/f/f21c0c70b37768f3e80e9ee909ef34c57635d5.png[/img]
Triangle $ABC$. Let $B_1$ and $C_1$ be such points, that $AB= BB_1, AC=CC_1$ and $B_1, C_1$ lie on the circumscribed circle $\Gamma$ of $\vartriangle ABC$. Perpendiculars drawn from from points $B_1$ and $C_1$ on the lines $AB$ and $AC$ intersect $\Gamma$ at points $B_2$ and $C_2$ respectively, these points lie on smaller arcs $AB$ and $AC$ of circle $\Gamma$ respectively, Prove that $BB_2 \parallel CC_2$.
On the equal $ AC $ and $ BC $ of an isosceles right triangle $ ABC $ , points $ D $ and $ E $ are marked respectively, so that $ CD = CE $. Perpendiculars on the straight line $ AE $, passing through the points $ C $ and $ D $, intersect the side $ AB $ at the points $ P $ and $ Q $.Prove that $ BP = PQ $.
In triangle $ABC$ point $M$ is the midpoint of side $AB$, and point $D$ is the foot of altitude $CD$.
Prove that $\angle A = 2\angle B$ if and only if $AC = 2MD$
An acute triangle $ABC$ is inscribed in circle $\omega$ centered at $O$. Line $BO$ and side $AC$ meet at $B_1$. Line $CO$ and side $AB$ meet at $C_1$. Line $B_1C_1$ meets circle $\omega$ at $P$ and $Q$. If $AP = AQ$, prove that $AB = AC$.
(i) $ABC$ is a triangle with a right angle at $A$, and $P$ is a point on the hypotenuse $BC$.
The line $AP$ produced beyond $P$ meets the line through $B$ which is perpendicular to $BC$ at $U$.
Prove that $BU = BA$ if, and only if, $CP = CA$.
(ii) $A$ is a point on the semicircle $CB$, and points $X$ and $Y$ are on the line segment $BC$.
The line $AX$, produced beyond $X$, meets the line through $B$ which is perpendicular to $BC$ at $U$.
Also the line $AY$, produced beyond $Y$, meets the line through $C$ which is perpendicular to $BC$ at $V$.
Given that $BY = BA$ and $CX = CA$, determine the angle $\angle VAU$.
Let $ ABC$ be an acute-angled, non-isosceles triangle with orthocenter $H$, $M$ midpoint of side $AB$ and $w$ bisector of angle $\angle ACB$. Let $S$ be the point of intersection of the perpendicular bisector of side $AB$ with $w$ and $F$ the foot of the perpendicular from $H$ on $w$. Prove that the segments $MS$ and $MF$ are equal.
[i](Karl Czakler)[/i]
The median $AM$ is drawn in the triangle $ABC$ ($AB \ne AC$). The point $P$ is the foot of the perpendicular drawn on the segment $AM$ from the point $B$. On the segment $AM$ we chose such a point $Q$ that $AQ = 2PM$. Prove that $\angle CQM = \angle BAM$.
Let $ABC$ be a triangle, $P$ the midpoint of $BC$, and $Q$ a point on segment $CA$ such that $|CQ| = 2|QA|$. Let $S$ be the intersection of $BQ$ and $AP$. Prove that $|AS| = |SP|$.
Consider the pentagon $ABCDE$ such that $AB = AE = x$, $AC = AD = y$, $\angle BAE = 90^o$ and $\angle ACB = \angle ADE = 135^o$. It is known that $C$ and $D$ are inside the triangle $BAE$. Determine the length of $CD$ in terms of $x$ and $y$.
The point $P$ lies on the edge $AB$ of a quadrilateral $ABCD$.
The angles $BAD, ABC$ and $CPD$ are right, and $AB = BC + AD$.
Show that $BC = BP$ or $AD = BP$.
Point $D$ on the side $BC$ of the acute triangle $ABC$ is chosen so that $AD = AC$. Let $P$ and $Q$ be the feet of the perpendiculars from $C$ and $D$ on the side $AB$, respectively. Suppose that $AP^2 + 3BP^2 = AQ^2 + 3BQ^2$. Determine the measure of angle $\angle ABC$.
Let $O$ be the center of the circle circumscribed to $\vartriangle ABC$, and let $ P$ be any point on $BC$ ($P \ne B$ and $P \ne C$). Suppose that the circle circumscribed to $\vartriangle BPO$ intersects $AB$ at $R$ ($R \ne A$ and $R \ne B$) and that the circle circumscribed to $\vartriangle COP$ intersects $CA$ at point $Q$ ($Q \ne C$ and $Q \ne A$).
1) Show that $\vartriangle PQR \sim \vartriangle ABC$ and that$ O$ is orthocenter of $\vartriangle PQR$.
2) Show that the circles circumscribed to the triangles $\vartriangle BPO$, $\vartriangle COP$, and $\vartriangle PQR$ all have the same radius.