Found problems: 361
Let $ABCD$ be a cyclic quadrilateral with $\angle ADC = \angle DBA$. Furthermore, let $E$ be the projection of $A$ on $BD$. Show that $BC = DE - BE$ .
Let $ABC$ be an isosceles triangle with $AB = AC$. Let P be a point in the interior of $ABC$ such that $PB > PC$ and $\angle PBA = \angle PCB$. Let $M$ be the midpoint of the side $BC$. Let $O$ be the circumcenter of the triangle $APM$. Prove that $\angle OAC=2 \angle BPM$ .
On the diagonal $AC$ of a convex quadrilateral $ABCD$ is chosen such a point $K$ such that $KD = DC$, $\angle BAC = \frac12 \angle KDC$, $\angle DAC = \frac12 \angle KBC$. Prove that $\angle KDA = \angle BCA$ or $\angle KDA = \angle KBA$.
Let $ABCD$ be a parallelogram which is not a rectangle and $E$ be the point in its plane such that $AE \perp AB$ and $CE \perp CB$. Prove that $\angle DEA = \angle CEB$.
Let $ABC$ be a triangle with $\angle BAC= 45^o$ . Altitudes $AD, BE, CF$ meet at $H$. $EF$ cuts $BC$ at $P$. $I$ is the midpoint of $BC$, $IF$ cuts $PH$ in $Q$.
a) Prove that $\angle IQH = \angle AIE$.
b) Let $(K)$ be the circumcircle of triangle $ABC$, $(J)$ be the circumcircle of triangle $KPD$. $CK$ cuts circle $(J)$ at $G$, $IG$ cuts $(J)$ at $M$, $JC$ cuts circle of diameter $BC$ at $N$. Prove that $G, N, M, C$ lie on the same circle.
In the acute-angled triangle $ABC$ with circumcircle $\omega$ and orthocenter $H$, points $D$ and $E$ are the feet of the perpendiculars from $A$ onto $BC$ and from $B$ onto $AC$ respecively. Let $P$ be a point on the minor arc $BC$ of $\omega$ . Points $M$ and $N$ are the feet of the perpendiculars from $P$ onto lines $BC$ and $AC$ respectively. Let $PH$ and $MN$ intersect at $R$. Prove that $\angle DMR=\angle MDR$.
A convex pentagon $ABCDE$ with $DC = DE$ and $\angle DCB = \angle DEA = 90^o$ is given.
Let $F$ be a point on the segment $AB$ such that $AF : BF = AE : BC$.
Prove that $\angle FCE = \angle ADE$ and $\angle FEC = \angle BDC$.
On the sides $AB, \, \, BC, \, \, CA$ of the triangle $ABC$ the points ${{C} _ {1}}, \, \, {{A} _ { 1}},\, \, {{B} _ {1}}$ are selected respectively, that are different from the vertices. It turned out that $\Delta {{A} _ {1}} {{B} _ {1}} {{C} _ {1}}$ is equilateral, $\angle B{{C}_{1}}{{A}_{1}}=\angle {{C}_{1}}{{B}_{1}}A$ and $\angle B{{A}_{1}}{{C}_{1}}=\angle {{A}_{1}}{{B}_{1}}C$ . Is $ \Delta ABC$ equilateral?
$ABC$ is a triangle with $\angle A > 90^o$ . On the side $BC$, two distinct points $P$ and $Q$ are chosen such that $\angle BAP = \angle PAQ$ and $BP \cdot CQ = BC \cdot PQ$. Calculate the size of $\angle PAC$.
Let $ H $ be the point of intersection of the altitudes $ AP $ and $ CQ $ of the acute-angled triangle $ABC$. The points $ E $ and $ F $ are marked on the median $ BM $ such that $ \angle APE = \angle BAC $, $ \angle CQF = \angle BCA $, with point $ E $ lying inside the triangle $APB$ and point $ F $ is inside the triangle $CQB$. Prove that the lines $AE, CF$, and $BH$ intersect at one point.
In triangle $ABC$, angle $C$ is a right angle. Found on the side $AC$ point $D$, and on the segment $BD$, point $K$ such that $\angle ABC = \angle KAD =\angle AKD$. Prove that $BK = 2DC$.
Polygon $A_0A_1...A_{n-1}$ satisfies the following:
$\bullet$ $A_0A_1 \le A_1A_2 \le ...\le A_{n-1}A_0$ and
$\bullet$ $\angle A_0A_1A_2 = \angle A_1A_2A_3 = ... = \angle A_{n-2}A_{n-1}A_0$ (all angles are internal angles).
Prove that this polygon is regular.
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)
Given $\vartriangle ABC$ with circumcenter $O$. Let $D$ be a point satisfying $\angle ABD = \angle DCA$ and $M$ be the midpoint of $AD$. Suppose that $BM,CM$ intersect circle $(O)$ at another points $E, F$, respectively. Let $P$ be a point on $EF$ so that $AP$ is tangent to circle $(O)$. Prove that $A, P,M,O$ are concyclic.
[img]https://2.bp.blogspot.com/-gSgUG6oywAU/XnSKTnH1yqI/AAAAAAAALdw/3NuPFuouCUMO_6KbydE-KIt6gCJ4OgWdACK4BGAYYCw/s320/imoc2017%2Bg7.png[/img]
Let $P$ be an interior point of a circle and $A_1,A_2...,A_{10}$ be points on the circle such that $\angle A_1PA_2 = \angle A_2PA_3 = ... = \angle A_{10}PA_1 = 36^o$. Prove that $PA_1 + PA_3 + PA_5 + PA_7 +PA_9 = PA_2 + PA_4 + PA_6 + PA_8 + PA_{10}$.
Let $ABC$ be a triangle. Let $P_1$ and $P_2$ be points on the side $AB$ such that $P_2$ lies on the segment $BP_1$ and $AP_1 = BP_2$; similarly, let $Q_1$ and $Q_2$ be points on the side $BC$ such that $Q_2$ lies on the segment $BQ_1$ and $BQ_1 = CQ_2$. The segments $P_1Q_2$ and $P_2Q_1$ meet at $R$, and the circles $P_1P_2R$ and $Q_1Q_2R$ meet again at $S$, situated inside triangle $P_1Q_1R$. Finally, let $M$ be the midpoint of the side $AC$. Prove that the angles $P_1RS$ and $Q_1RM$ are equal.
On the sides $AB$ and $AD$ of the square $ABCD$, the points $N$ and $P$ are selected respectively such that $NC=NP$. The point $Q$ is chosen on the segment $AN$ so that $\angle QPN = \angle NCB$. Prove that $2\angle BCQ = \angle AQP$.
Consider acute triangles $ABC$ and $BCD$, with $\angle BAC = \angle BDC$, such that $A$ and $D$ are on opposite sides of line $BC$. Denote by $E$ the foot of the perpendicular line to $AC$ through $B$ and by $F$ the foot of the perpendicular line to $BD$ through $C$. Let $H_1$ be the orthocenter of triangle $ABC$ and $H_2$ be the orthocenter of $BCD$. Show that lines $AD, EF$ and $H_1H_2$ are concurrent.
The midpoint of the hypotenuse $AB$ of the right triangle $ABC$ is $K$. The point $M$ on the side $BC$ is taken such that $BM = 2 \cdot MC$. Prove that $\angle BAM = \angle CKM$.
Consider a triangle $ABC$ and let $M$ be the midpoint of the side $BC$.
Suppose $\angle MAC = \angle ABC$ and $\angle BAM = 105^o$. Find the measure of $\angle ABC$.
The points $C$ and $D$ lie on a halfline from the midpoint $M$ of a segment $AB$, so that $|AC| = |BD|$. Prove that the angles $u = \angle ACM$ and $v = \angle BDM$ are equal.
[img]https://1.bp.blogspot.com/-tQEJ1VBCa8U/XzT7IhwlZHI/AAAAAAAAMVI/xpRdlj5Rl64VUt_tCRsQ1UxIsv_SGrMlACLcBGAsYHQ/s0/2014%2BMohr%2Bp3.png[/img]
On the sides $AB, BC$ and $CA$ of a triangle $ABC$ are located the points $P, Q$ and $R$ respectively, such that $BQ = 2QC, CR = 2RA$ and $\angle PRQ = 90^o$. Show that $\angle APR =\angle RPQ$.
In the triangle $ABC$, $AA_1$ and $BB_1$ are altitudes. On the side $AB$ , points $M$ and $K$ are selected so that $B_1K \parallel BC$ and $A_1M \parallel AC$. Prove that the angle $AA_1K$ is equal to the angle $BB_1M$.
(D. Prokopenko)
Let $ABC$ be an acute triangle and let $O$ be its circumcentre. Now, let the diameter $PQ$ of circle $ABC$ intersects sides $AB$ and $AC$ in their interior at$ D$ and $E$, respectively. Now, let $F$ and $G$ be the midpoints of $CD$ and $BE$. Prove that $\angle FOG=\angle BAC$
A line meets a segment $AB$ at point $C$. Which is the maximal number of points $X$ of this line such that one of angles $AXC$ and $BXC$ is equlal to a half of the second one?