Found problems: 131
In an acute-angled triangle $ABC$, point $I$ is the center of the inscribed circle, point $T$ is the midpoint of the arc $ABC$ of the circumcircle of triangle $ABC$. It turned out that $\angle AIT = 90^o$ . Prove that $AB + AC = 3BC$.
(Matthew of Kursk)
The center of square $ABCD$ is $K$. The point $P$ is chosen such that $P \ne K$ and the angle $\angle APB$ is right . Prove that the line $PK$ bisects the angle between the lines $AP$ and $BP$.
Consider a parallelogram $ABCD$ such that the midpoint $M$ of the side $CD$ lies on the angle bisector of $\angle BAD$. Show that $\angle AMB$ is a right angle.
Given the pyramid $ABCD$. Let $O$ be the midpoint of edge $AC$. Given that $DO$ is the height of the pyramid, $AB=BC=2DO$ and the angle $ABC$ is right. Cut this pyramid into $8$ equal and similar to it pyramids.
Let $ABCD$ be a trapezoid, with $AD \parallel BC$, let $M$ be the midpoint of $AD$, and let $C_1$ be symmetric point to $C$ with respect to line $BD$. Segment $BM$ meets diagonal $AC$ at point $K$, and ray $C_1K$ meets line $BD$ at point $H$. Prove that $\angle{AHD}$ is a right angle.
[i]Proposed by Giorgi Arabidze, Georgia[/i]
On the sides $AB, BC$ and $AC$ of the triangle $ABC$, points $C', A'$ and $B'$ are selected, respectively, so that the angle $A'C'B'$ is right. Prove that the segment $A'B'$ is longer than the diameter of the inscribed circle of the triangle $ABC$.
(M. Volchkevich)
In the right trapezoid $ABCD$ with $AB \parallel CD, \angle B = 90^o$ and $AB = 2DC$.
At points $A$ and $D$ there is therefore a part of the plane $(ABC)$ perpendicular to the plane of the trapezoid, on which the points $N$ and $P$ are taken, ($AP$ and $PD$ are perpendicular to the plane) such that $DN = a$ and $AP = \frac{a}{2}$ . Knowing that $M$ is the midpoint of the side $BC$ and the triangle $MNP$ is equilateral, determine:
a) the cosine of the angle between the planes $MNP$ and $ABC$.
b) the distance from $D$ to the plane $MNP$
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 $C$ be a right angle in triangle $ABC$. On legs $AC$ and$BC$ the squares $ACKL, BCMN$ are constructed outside of triangle. If $CE$ is an altitude of the triangle prove that $LEM$ is a right angle.
The diagonals of the trapezoid $ABCD$ are perpendicular ($AD//BC, AD>BC$) . Point $M$ is the midpoint of the side of $AB$, the point $N$ is symmetric of the center of the circumscribed circle of the triangle $ABD$ wrt $AD$. Prove that $\angle CMN = 90^o$.
(A. Mudgal, India)
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]
On the trapezoid $ABCD$ , side $DA$ is perpendicular to the bases $AB$ and $CD$ . The base $AB$ measures $45$, the base $CD$ measures $20$ and the $BC$ side measures $65$. Let $P$ on the $BC$ side such that $BP$ measures $45$ and $M$ is the midpoint of $DA$. Calculate the measure of the $PM$ segment.
Given is a triangle $ABC$ and the points $M, P$ lie on the segments $AB, BC$, respectively, such that $AM=BC$ and $CP=BM$. If $AP$ and $CM$ meet at $O$ and $2\angle AOM=\angle ABC$, find the measure of $\angle ABC$.
In the rectangle $ABCD$, point $M$ is the midpoint of the side $BC$. The points $P$ and $Q$ lie on the diagonal $AC$ such that $\angle DPC = \angle DQM = 90^o$. Prove that $Q$ is the midpoint of the segment $AP$.
In triangle $ABC$, on line $CA$ it is given point $D$ such that $CD = 3 \cdot CA$ (point $A$ is between points $C$ and $D$), and on line $BC$ it is given point $E$ ($E \neq B$) such that $CE=BC$. If $BD=AE$, prove that $\angle BAC= 90^{\circ}$
In an acute triangle $AB$ the heights $ BE$ and $CF$ intersect at the orthocenter $H$, and $M$ is the midpoint of $BC$. The line $EF$ intersects the lines $MH$ and $BC$ at the points $P$ and $T$ , respectively. $AP$ intersects the cirumcscribed circle of $\vartriangle ABC$ for second time at the point $Q$ . Prove that $\angle AQT= 90^o$.
(Fedir Yudin)
In a circle $S$, a chord $AB$ is drawn and let $M$ be the midpoint of the arc $AB$. Let $P$ be a point in segment $AB$ other than its midpoint. The extension of the segment $MP$ cuts $S$ in $Q$. Let $S_1$ be the circle that is tangent to the AP segments and $MP$, and also is tangent to $S$, and let $S_2$ be the circle that is tangent to the segments $BP$ and $MP$, and also tangent to $S$. The common outer tangent lines to the circles $S_1$ and $S_2$ are cut at $C$. Prove that $\angle MQC = 90^o$.
Let $ABC$ be a triangle with $AB\ne AC$. Its incircle has center $I$ and touches the side $BC$ at point $D$. Line $AI$ intersects the circumcircle $\omega$ of triangle $ABC$ at $M$ and $DM$ intersects again $\omega$ at $P$. Prove that $\angle API= 90^o$.
In triangle $ABC,\angle B > \angle C, T$ is the midpoint of arc $BAC$ of the circumcicle of $ABC$, and $I$ is the incentre of $ABC$. Let $E$ be point such that $\angle AEI = 90^0$ and $AE$ is parallel to $BC$. If $TE$ intersects the circumcircle of $ABC$ at $P(\ne T)$ and $\angle B = \angle IPB$, determine $\angle A$.
Let $ABCDE$ be a pentagon with right angles at vertices $B$ and $E$ and such that $AB = AE$ and $BC = CD = DE$. The diagonals $BD$ and $CE$ meet at point $F$. Prove that $FA = AB$.
In the cyclic quadrilateral $ABCD$, the sides $AB, DC$ meet at $Q$, the sides $AD,BC$ meet at $P, M$ is the midpoint of $BD$, If $\angle APQ=90^o$, prove that $PM$ is perpendicular to $AB$.
Let $ABC$ be a triangle with two angles $B,C$ not having the same measure, $I$ be its incircle, $(O)$ its circumcircle. Circle $(O_b)$ touches $BA,BC$ and is internally tangent to $(O)$ at $B_1$. Circle $(O_c)$ touches $CA,CB$ and is internally tangent to $(O)$ at $C_1$. Let $S$ be the intersection of $BC$ and $B_1C_1$. Prove that $\angle AIS = 90^o$.
Nguyễn Minh Hà
Point $C$ lies inside the right angle $AOB$. Prove that the perimeter of triangle $ABC$ is greater than $2 OC$.
A tetrahedron $ABCD$ satisfies $\angle BAC=\angle CAD=\angle DAB=90^o$. Show that the areas of its faces satisfy the equation $area(BAC)^2 + area(CAD)^2 + area(DAB)^2 = area(BCD)^2$.
.
Given is a triangle $ABC$ and the points $M, P$ lie on the segments $AB, BC$, respectively, such that $AM=BC$ and $CP=BM$. If $AP$ and $CM$ meet at $O$ and $2\angle AOM=\angle ABC$, find the measure of $\angle ABC$.