Found problems: 361
Given a triangle $ABC$, in which the angle $B$ is three times the angle $C$. On the side $AC$, point $D$ is chosen such that the angle $BDC$ is twice the angle $C$. Prove that $BD + BA = AC$.
In the isosceles triangle $ABC$ we have $AC = BC$. Let $X$ be an arbitrary point of the segment $AB$. The line parallel to $BC$ and passing through $X$ intersects the segment $AC$ in $N$, and the line parallel to $AC$ and passing through $BC$ intersects the segment $BC$ in $M$. Let $k_1$ be the circle with center $N$ and radius $NA$. Similarly, let $k_2$ be the circle with center $M$ and radius $MB$. Let $T$ be the intersection of the circles $k_1$ and $k_2$ different from $X$. Show that the angles $\angle NCM$ and $\angle NTM$ are equal.
Consider triangle $ABC$ such that $AB \le AC$. Point $D$ on the arc $BC$ of thecircumcirle of $ABC$ not containing point $A$ and point $E$ on side $BC$ are such that $\angle BAD = \angle CAE < \frac12 \angle BAC$ . Let $S$ be the midpoint of segment $AD$. If $\angle ADE = \angle ABC - \angle ACB$ prove that $\angle BSC = 2 \angle BAC$ .
Let $AD$ and $AE$ be the altitude and median of triangle $ABC$, in with $\angle B = 2\angle C$. Prove that $AB = 2DE$.
Let $ABC$ be a triangle and let $O$ be its circumcentre. The internal and external bisectrices of the angle $BAC$ meet the line $BC$ at points $D$ and $E$, respectively. Let further $M$ and $L$ respectively denote the midpoints of the segments $BC$ and $DE$. The circles $ABC$ and $ALO$ meet again at point $N$. Show that the angles $BAN$ and $CAM$ are equal.
Let $ABCD$ be a cyclic quadrilateral satisfying $\angle ABD = \angle DBC$. Let $E$ be the intersection of the diagonals $AC$ and $BD$. Let $M$ be the midpoint of $AE$, and $N$ be the midpoint of $DC$. Show that $MBCN$ is a cyclic quadrilateral.
In the quadrilateral $ABCD$, shown in fig. , the equations are true: $\angle ABC = \angle BCD$ and $2AB = CD$. On the side $BC$, a point $X$ is selected such that $\angle BAX = \angle CDA$. Prove that $AX = AD$.
[img]https://cdn.artofproblemsolving.com/attachments/2/9/0884eb311d1e40300c1e5980fd53eaadfa7a25.png[/img]
Let $ABCDE$ be a convex pentagon such that angles $CAB$, $BCA$, $ECD$, $DEC$ and $AEC$ are equal. Prove that $CE$ bisects $BD$.
Let ${\omega}$ be the circumcircle of an acute-angled triangle ${ABC}$. Point ${D}$ lies on the arc ${BC}$ of ${\omega}$ not containing point ${A}$. Point ${E}$ lies in the interior of the triangle ${ABC}$, does not lie on the line ${AD}$, and satis fies ${\angle DBE =\angle ACB}$ and ${\angle DCE = \angle ABC}$. Let ${F}$ be a point on the line ${AD}$ such that lines ${EF}$ and ${BC}$ are parallel, and let ${G}$ be a point on ${\omega}$ different from ${A}$ such that ${AF = FG}$. Prove that points ${D,E, F,G}$ lie on one circle.
(Slovakia)
In the acute triangle $ ABC $, the point $ I $ is the center of the inscribed the circle, the point $ O $ is the center of the circumscribed circle and the point $ I_a $ is the center the excircle tangent to the side $ BC $ and the extensions of the sides $ AB $ and $ AC $. Point $ A'$ is symmetric to vertex $ A $ with respect to the line $ BC $. Prove that $ \angle IOI_a = \angle IA'I_a $.
In the convex quadrilateral $ABCD$ we have that $\angle BCD = \angle ADC \ge 90 ^o$. The bisectors of $\angle BAD$ and $\angle ABC$ intersect in $M$. Prove that if $M \in CD$, then $M$ is the middle of $CD$.
Two trapezoid angles and diagonals are respectively equal. Is it true that such are the trapezoid equal?
Given an isosceles triangle $ABC$ with base $BC$. Inside the side $BC$ is given a point $D$. Let $E, F$ be respectively points on the sides $AB, AC$ that $|\angle BED | = |\angle DF C| > 90^o$ . Prove that the circles circumscribed around the triangles $ABF$ and $AEC$ intersect on the line $AD$ at a point different from point $A$.
(Patrik Bak, Michal Rolínek)
In the triangle $[ABC], D$ is the midpoint of $[AB]$ and $E$ is the trisection point of $[BC]$ closer to $C$. If $\angle ADC= \angle BAE$ , find the measue of $\angle BAC$ .
Let $ABCM$ be a quadrilateral and $D$ be an interior point such that $ABCD$ is a parallelogram. It is known that $\angle AMB =\angle CMD$. Prove that $\angle MAD =\angle MCD$.
Given a convex pentagon $ABCDE$ with $\angle ABC= \angle ADE$ and $\angle AEC= \angle ADB$ . Prove that $\angle BAC = \angle DAE$ .
Let $E$ be a point lies inside the parallelogram $ABCD$ such that $\angle BCE = \angle BAE$.
Prove that the circumcenters of triangles $ABE,BCE,CDE,DAE$ are concyclic.
Let $ABC$ be an acute-angled scalene triangle and $T$ be a point inside it such that $\angle ATB = \angle BTC = 120^o$. A circle centered at point $E$ passes through the midpoints of the sides of $ABC$. For $B, T, E$ collinear, find angle $ABC$.
Given a triangle $ABC$. Construct a point $D$ on the side $AB$ and point $E$ on the side $AC$ so that $BD = CE$ and $\angle ADC = \angle BEC$
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]
Given a convex pentagon $ABCDE$, with $\angle BAC = \angle ABE = \angle DEA - 90^o$, $\angle BCA = \angle ADE$ and also $BC = ED$. Prove that $BCDE$ is parallelogram.
Let $ABC$ be a triangle and $D, E, F$ the midpoints of the sides $BC, CA, AB$ respectively. Show that $\angle DAC = \angle ABE$ if and only if $\angle AFC = \angle BDA$
In the triangle $ABC$, ${{A}_{1}}$ and ${{C}_{1}} $ are the midpoints of sides $BC $ and $AB$ respectively. Point $P$ lies inside the triangle. Let $\angle BP {{C}_{1}} = \angle PCA$. Prove that $\angle BP {{A}_{1}} = \angle PAC $.
Let $O$ be the circumcenter and $H$ the orthocenter of an acute-triangle $ABC$. The perpendicular bisector of $AO$ intersects the line $BC$ at point $S$. Let $L$ be the midpoint of $OH$. Prove that $\angle OAH = \angle LSA$.
As shown in the figure, in the parallelogram $ABCD$, $I$ is the incenter of $\vartriangle BCD$, and $H$ is the orthocenter of $\vartriangle IBD$. Prove that $\angle HAB=\angle HAD$.
[img]https://cdn.artofproblemsolving.com/attachments/4/3/5fa16c208ef3940443854756ae7bdb9c4272ed.png[/img]