Found problems: 131
In the convex quadrilateral of the $ABCD$, the diagonals of $AC$ and $BD$ are mutually perpendicular and intersect at point $E$. On the side of $AD$, a point $P$ is chosen such that $PE = EC$. The circumscribed circle of the triangle $BCD$ intersects the segment $AD$ at the point $Q$. The circle passing through point $A$ and tangent to the line $EP$ at point $P$ intersects the segment $AC$ at point $R$. It turns out that points $B, Q, R$ are collinear. Prove that $\angle BCD = 90^o$.
In a convex quadrilateral $ABCD$, $\angle ABC = 90^o$ , $\angle BAC = \angle CAD$, $AC = AD, DH$ is the alltitude of the triangle $ACD$. In what ratio does the line $BH$ divide the segment $CD$?
Let $\vartriangle ABC$ be a triangle. The angle bisector of $\angle CAB$ intersects$ BC$ at $L$. On the interior of line segments $AC$ and $AB$, two points, $M$ and $N$, respectively, are chosen in such a way that the lines $AL, BM$ and $CN$ are concurrent, and such that $\angle AMN = \angle ALB$. Prove that $\angle NML = 90^o$.
Given triangle $ABC$. Point $A_1$ is symmetric to vertex $A$ wrt line $BC$, and point $C_1$ is symmetric to vertex $C$ wrt line $AB$. Prove that if points $A_1$, $B$ and $C_1$ lie on the same line and $C_1B = 2A_1B$, then angle $\angle CA_1B$ is right.
Prove that all quadrilaterals $ABCD$ where $\angle B = \angle D = 90^o$, $|AB| = |BC|$ and $|AD| + |DC| = 1$, have the same area.
[img]https://1.bp.blogspot.com/-55lHuAKYEtI/XzRzDdRGDPI/AAAAAAAAMUk/n8lYt3fzFaAB410PQI4nMEz7cSSrfHEgQCLcBGAsYHQ/s0/2016%2Bmohr%2Bp3.png[/img]
Let $C, B, E$ be three points on a straight line $\ell$ in that order. Suppose that $A$ and $D$ are two points on the same side of $\ell$ such that
(i) $\angle ACE = \angle CDE = 90^o$ and
(ii) $CA = CB = CD$.
Let $F$ be the point of intersection of the segment $AB$ and the circumcircle of $\vartriangle ADC$.
Prove that $F$ is the incentre of $\vartriangle CDE$.
A quadrilateral $ABCD$ has a right angle at $\angle ABC$ and satisfies $AB = 12$, $BC = 9$, $CD = 20$, and $DA = 25$. Determine $BD^2$.
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Let $ABC$ be a triangle with circumcircle $\Gamma$. Let $\omega_0$ be a circle tangent to chord $AB$ and arc $ACB$. For each $i = 1, 2$, let $\omega_i$ be a circle tangent to $AB$ at $T_i$ , to $\omega_0$ at $S_i$ , and to arc $ACB$. Suppose $\omega_1 \ne \omega_2$. Prove that there is a circle passing through $S_1, S_2, T_1$, and $T_2$, and tangent to $\Gamma$ if and only if $\angle ACB = 90^o$.
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Let $H$ be the orthocenter of acute-angled $\vartriangle ABC$, and $X, Y$ points on the ray $AB, AC$. ($B$ lies between $X, A$, and $C$ lies between $Y, A$.) Lines $HX, HY$ intersect $BC$ at $D, E$ respectively. Let the line through $D$ parallel to $AC$ intersect $XY$ at $Z$. Prove that $\angle XHY = 90^o$ if and only if $ZE \parallel AB$.
On sides $AB$ and $BC$ of a square $ABCD$ the respective points $E$ and $F$ have been chosen so that $BE = BF$. Let $BN$ be the altitude in triangle $BCE$. Prove that $\angle DNF = 90$.
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 $AD = BP$.
Let a triangle $ABC, A_1$ be the midpoint of the segment $[BC], B_1 \in (AC)$ ¸and $C_1 \in (AB)$ such that $[A_1B_1$ is the bisector of the angle $AA_1C$ and $A_1C_1$ is perpendicular to $AB$. Show that the lines $AA_1, BB_1$ and $CC_1$ are concurrent if and only if $ \angle BAC = 90^o$
The center of a circle, the radius of which is $r$, lies on the bisector of the right angle $A$ at a distance $a$ from its sides ($a > r$). A tangent to the circle intersects the sides of the angle at points $B$ and $C$. Find the smallest possible value of the area of triangle $ABC$.
The base $AB$ of a trapezium $ABCD$ is longer than the base $CD$, and $\angle ADC$ is a right angle. The diagonals $AC$ and $BD$ are perpendicular. Let $E$ be the foot of the altitude from $D$ to the line $BC$. Prove that
$$\frac{AE}{BE} =\frac{ AC \cdot CD}{AC^2 - CD^2}$$
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Let $ABCD$ be a parallelogram and $P$ be an arbitrary point of the circumcircle of $\Delta ABD$, different from the vertices. Line $PA$ intersects the line $CD$ at point $Q$. Let $O$ be the center of the circumcircle $\Delta PCQ$. Prove that $\angle ADO = 90^o$.
In the quadrilateral $ABCD $, $AB = BC $, the point $K $ is the midpoint of the side $CD $, the rays $BK $ and $AD $ intersect at the point $M $ , the circumscribed circle $ \Delta ABM $ intersects the line $AC $ for the second time at the point $P $. Prove that $\angle BKP = 90 {} ^ \circ $.
(Anton Trygub)
Points $K$ and $L$ are on side $AB$ of triangle $ABC$ such that $KL=BC$ and $AK=LB$. Let $M$ be a midpoint of $AC$. Prove that $\angle KML = 90^{\circ}$
In a cyclic quadrilateral $ABCD$, let the diagonals $AC$ and $BD$ intersect at $X$. Let the circumcircles of triangles $AXD$ and $BXC$ intersect again at $Y$ . If $X$ is the incentre of triangle $ABY$ , show that $\angle CAD = 90^o$.
A $6$ meter ladder rests against a vertical wall. The midpoint of the ladder is twice as far from the ground as it is from the wall. At what height on the wall does the ladder reach?
Points $P$ and $Q$ lies inside triangle $ABC$ such that $\angle ACP =\angle BCQ$ and $\angle CAP = \angle BAQ$. Denote by $D,E$, and $F$ the feet of perpendiculars from $P$ to lines $BC,CA$, and $AB$, respectively. Prove that if $\angle DEF = 90^o$, then $Q$ is the orthocenter of triangle $BDF$.
Given a right triangle $ABC$ ($\angle A <45^o$,$ \angle C = 90^o$), on the sides $AC$ and $AB$ which are selected points $D,E$ respectively, such that $BD = AD$ and $CB = CE$. Let the segments $BD$ and $CE$ intersect at the point $O$. Prove that $\angle DOE = 90^o$.
Let $ABC $be a right triangle with $\angle C = 90^o$ and let $D$ be the foot of the altitude from $C$. Let $E$ be the centroid of triangle $ACD$ and let $F$ be the centroid of triangle $BCD$. The point $P$ satisfies $\angle CEP = 90^o$ and $|CP| = |AP|$, while point $Q$ satisfies $\angle CFQ = 90^o$ and $|CQ| = |BQ|$.
Prove that $PQ$ passes through the centroid of triangle $ABC$.
Given a triangle $ABC$ such that $AB<AC$ . On sides $AB$ and $BC$, points $P$ and $Q$ are marked respectively such that the lines $AQ$ and $CP$ are perpendicular, and the circle inscribed in the triangle $ABC$ touches the length $PQ$. The line $CP$ intersects the circle circumscribed around the triangle $ABC$ at the points $C$ and $T$. If the lines $CA,PQ$ and $BT$ intersect at one point, prove that the angle $\angle CAB$ is right.
Let $G, H$ be the centroid and orthocentre of $\vartriangle ABC$ which has an obtuse angle at $\angle B$. Let $\omega$ be the circle with diameter $AG$. $\omega$ intersects $\odot(ABC)$ again at $L \ne A$. The tangent to $\omega$ at $L$ intersects $\odot(ABC)$ at $K \ne L$. Given that $AG = GH$, prove $\angle HKG = 90^o$
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[i]Sam Bealing, United Kingdom[/i]
Let $ABC$ be a triangle. Four distinct points $D,A,B,E$ lie on the line $AB$ in this order such that $DA=AB=BE.$ Find necessary and sufficient condition for lengths $a=BC,b=AC$ such that the angle $\angle DCE$ is right.