Found problems: 721
(a) A circle is inscribed in a triangle. The tangent points are the vertices of a second triangle in which another circle is inscribed. Its tangency points are the vertices of a third triangle. The angles of this triangle are identical to those of the first triangle. Find these angles.
(b) A circle is inscribed in a scalene triangle. The tangent points are vertices of another triangle, in which a circle is inscribed whose tangent points are vertices of a third triangle, in which a third circle is inscribed, etc. Prove that the resulting sequence does not contain a pair of similar triangles.
Let $ \,ABC\,$ be a triangle and $ \,P\,$ an interior point of $ \,ABC\,$. Show that at least one of the angles $ \,\angle PAB,\;\angle PBC,\;\angle PCA\,$ is less than or equal to $ 30^{\circ }$.
Let the isosceles right triangle $ABC$ with $\angle A= 90^o$ . The points $E$ and $F$ are taken on the ray $AC$ so that $\angle ABE = 15^o$ and $CE = CF$. Determine the measure of the angle $CBF$.
In a convex quadrilateral $ABCD$ we are given that $\angle CBD = 10^o$, $\angle CAD = 20^o$, $\angle ABD = 40^o$, $\angle BAC = 50^o$. Determine the angles $\angle BCD$ and $\angle ADC$.
A square is drawn on a sheet of grid paper on the sides of the cells $ABCD$ with side $8$. Point $E$ is the midpoint of side $BC$, $Q$ is such a point on the diagonal $AC$ such that $AQ: QC = 3: 1$. Find the angle between straight lines $AE$ and $DQ$.
In an isosceles right triangle $ABC$ with a right angle $C$, point $M$ is the midpoint of leg $AC$. At the perpendicular bisector of $AC$, point $D$ was chosen such that $\angle CDM = 30^o$, and $D$ and $B$ lie on different sides of $AC$. Find the angle $\angle ABD$.
(Volodymyr Petruk)
A circle with center $O$ passes through the vertices $A$ and $C$ of the triangle $ABC$ and intersects the segments $AB$ and $BC$ again at distinct points $K$ and $N$ respectively. Let $M$ be the point of intersection of the circumcircles of triangles $ABC$ and $KBN$ (apart from $B$). Prove that $\angle OMB=90^{\circ}$.
Angle bisectors $AD$ and $BE$ are drawn in triangle $ABC$. It turned out that $DE$ is the bisector of triangle $ADC$. Find the angle $BAC$.
The circle $\omega_2$ passing through the center $O$ of the circle $\omega_1$, is tangent to the circle $\omega_2$ at the point $A$. On the circle $\omega_2$, the point $C$ is taken so that the ray $AC$ intersects the circle $\omega_1$ for second time at point $D$, the ray $OC$ intersects the circle $\omega_1$ at point $E$ and the lines $DE$ and $AO$ are parallel. Find the size of the angle $DAE$.
In an acute triangle $ABC$, the angle bisector$AD$ and altitude $BE$ are drawn. Prove that angle $CED$ is greater than $45^o$.
Let $ABCD$ be a square and let point $F$ be any point on side $BC$. Let the line perpendicular to $DF$, that passes through $B$, intersect line $DC$ at $Q$. What is value of $\angle FQC$?
In the triangle $ABC$, one of the angles of which is equal to $48^o$, side lengths satisfy $(a-c)(a+c)^2+bc(a+c)=ab^2$. Express in degrees the measures of the other two angles of this triangle.
In the triangle $ABC$ it is known that $2AC=AB$ and $\angle A = 2\angle B$. In this triangle draw the angle bisector $AL$, and mark point $M$, the midpoint of the side $AB$. It turned out that $CL = ML$. Prove that $\angle B= 30^o$.
(Hilko Danilo)
In the plane are $10$ lines in general position, which means that no $2$ are parallel and no $3$ are concurrent. Where $2$ lines intersect, we measure the smaller of the two angles formed between them. What is the maximum value of the sum of the measures of these $45$ angles?
Bisector of one angle of triangle $ABC$ is equal to the bisector of its external angle at the same vertex (see figure). Find the difference between the other two angles of the triangle.
[img]https://cdn.artofproblemsolving.com/attachments/c/3/d2efeb65544c45a15acccab8db05c8314eb5f2.png[/img]
The two tangents to the incircle of a right-angled triangle $ABC$ that are perpendicular to the hypotenuse $AB$ intersect it at the points $P$ and $Q$. Find $\angle PCQ$.
(M Evdokimov,)
Find the angle $BCA$ in the quadrilateral of the figure.
[img]https://cdn.artofproblemsolving.com/attachments/0/2/974e23be54125cde8610a78254b59685833b5b.png[/img]
The measure of the angle $\angle A$ of the acute triangle $ABC$ is $60^o$, and $HI = HB$, where $I$ and $H$ are the incenter and the orthocenter of the triangle $ABC$. Find the measure of the angle $\angle B$.
In triangle $ABC$, we have $\angle ABC=\angle ACB=44^o$. Point $M$ is in its interior such that $\angle MBC=16^o$ and $\angle MCB=30^o$. Prove that $\angle MAC=\angle MBC$.
[i] (Andra Elena Mircea)[/i]
In a triangle $ABC$, $\angle B= 2\angle A \ne 90^o$ . The inner bisector of $B$ intersects the perpendicular bisector of $[AC]$ at a point $D$. Prove that $AB \parallel CD$.
In the triangle $ABC$, $D$ is the midpoint of $AB$, and $E$ is the point on the side $BC$, for which $CE = \frac13 BC$. It is known that $\angle ADC =\angle BAE$. Find $\angle BAC$.
Let $\Gamma$ be a semicircle with diameter $AB$. On this diameter is selected a point $C$, and on the semicircle are selected points $D$ and $E$ so that $E$ lies between $B$ and $D$. It turned out that $\angle ACD = \angle ECB$. The intersection point of the tangents to $\Gamma$ at points $D$ and $E$ is denoted by $F$. Prove that $\angle EFD=\angle ACD+ \angle ECB$.
Let $O$ be the center of the circle $\omega$, and let $A$ be a point inside this circle, different from $O$. Find all points $P$ on the circle $\omega$ for which the angle $\angle OPA$ acquires the greatest value.
On two sides of an angle, points $A, B$ are chosen. The midpoint $M$ of the segment $AB$ belongs to two lines such that one of them meets the sides of the angle at points $A_1, B_1$, and the other at points $A_2, B_2$. The lines $A_1B_2$ and $A_2B_1$ meet $AB$ at points $P$ and $Q$. Prove that $M$ is the midpoint of $PQ$.
Given an isosceles triangle $BAC$ with vertex angle $\angle BAC =20^o$. Construct an equilateral triangle $BDC$ such that $D,A$ are on the same side wrt $BC$. Construct an isosceles triangle $DEB$ with vertex angle $\angle EDB = 80^o$ and $C,E$ are on the different sides wrt $DB$. Prove that the triangle $AEC$ is isosceles at $E$.