Found problems: 509
Let $k$ be the incircle of the triangle $ABC$ with the center of the incircle $I$. The circle $k$ touches the sides $BC, CA$ and $AB$ in points $D, E$ and $F$. Let $G$ be the intersection of the straight line $AI$ and the circle $k$, which lies between $A$ and $I$. Assume $BE$ and $FG$ are parallel. Show that $BD = EF$.
Let $ABC$ be an acute triangle, and let $D$ be the foot of the altitude through $A$. On $AD$, there are distinct points $E$ and $F$ such that $|AE| = |BE|$ and $|AF| =|CF|$. A point$ T \ne D$ satis es $\angle BTE = \angle CTF = 90^o$. Show that $|TA|^2 =|TB| \cdot |TC|$.
Given is triangle $ABC$ with $AB > AC$. Circles $O_B$, $O_C$ are inscribed in angle $BAC$ with $O_B$ tangent to $AB$ at $B$ and $O_C$ tangent to $AC$ at $C$. Tangent to $O_B$ from $C$ different than $AC$ intersects $AB$ at $K$ and tangent to $O_C$ from $B$ different than $AB$ intersects $AC$ at $L$. Line $KL$ and the angle bisector of $BAC$ intersect $BC$ at points $P$ and $M$, respectively. Prove that $BP = CM$.
A triangle $ABC$ with circumcenter $U$ is given, so that $\angle CBA = 60^o$ and $\angle CBU = 45^o$ apply. The straight lines $BU$ and $AC$ intersect at point $D$. Prove that $AD = DU$.
(Karl Czakler)
Let $ABCDE$ be a convex pentagon with $AB = BC =CA$ and $CD = DE = EC$.
Let $T$ be the centroid of $\vartriangle ABC$, and $N$ be the midpoint of $AE$. Compute $\angle NT D$
In a quadrilateral $ABCD$ we have $\angle A = \angle C = 90^o$. Let $E$ be a point in the interior of $ABCD$. Let $M$ be the midpoint of $BE$. Prove that $\angle ADB = \angle EDC$ if and only if $|MA| = |MC|$.
Let $ABCD$ be a square with circumcircle $\Gamma_1$. Let $P$ be a point on the arc $AC$ that also contains $B$. A circle $\Gamma_2$ touches $\Gamma_1$ in $P$ and also touches the diagonal $AC$ in $Q$. Let $R$ be a point on $\Gamma_2$ such that the line $DR$ touches $\Gamma_2$. Proof that $|DR| = |DA|$.
Let $ABC$ be an isosceles triangle. Suppose that points $K$ and $L$ are chosen on lateral sides $AB$ and $AC$ respectively so that $AK = CL$ and $\angle ALK + \angle LKB = 60^o$. Prove that $KL = BC$.
Let $ABC$ be a triangle, $P$ the midpoint of $BC$, and $Q$ a point on segment $CA$ such that $|CQ| = 2|QA|$. Let $S$ be the intersection of $BQ$ and $AP$. Prove that $|AS| = |SP|$.
On the side $AB$ of the triangle $ABC$ mark the point $K$. The segment $CK$ intersects the median $AM$ at the point $F$. It is known that $AK = AF$. Find the ratio $MF: BK$.
Consider triangle $ABC$ with $AB = AC$ and $\angle A = 40 ^o$. The points $S$ and $T$ are on the sides $AB$ and $BC$, respectively, so that $\angle BAT = \angle BCS= 10 ^o$. The lines $AT$ and $CS$ intersect at point $P$. Prove that $BT = 2PT$.
Let $ABCD$ be a trapezoid with $\angle A = \angle B = 90^o$ and a point $E$ lies on the segment $CD$. Denote $(\omega)$ as incircle of triangle $ABE$ and it is tangent to $AB,AE,BE$ respectively at $P, F,K$. Suppose that $KF$ cuts $BC,AD$ at $M,N$ and $PM,PN$ cut $(\omega)$ at $H, T$. Prove that $PH = PT$.
Given a cyclic $ABCD$ with $AB=AD$. Points $M$ and $N$ are marked on the sides $CD$ and $BC$, respectively, so that $DM+BN=MN$. Prove that the circumcenter of the triangle $AMN$ belongs to the segment $AC$.
N.Sedrakian
Points $K$ and $M$ are taken on the sides $AB$ and $CD$ of square $ABCD$, respectively, and on the diagonal $AC$ - point $L$ such that $ML = KL$. Let $P$ be the intersection point of the segments $MK$ and $BD$. Find the angle $\angle KPL$.
In triangle $ABC, \angle A$ is smaller than $\angle C$. Point $D$ lies on the (extended) line $BC$ (with $B$ between $C$ and $D$) such that $|BD| = |AB|$. Point $E$ lies on the bisector of $\angle ABC$ such that $\angle BAE = \angle ACB$. Line segment $BE$ intersects line segment $AC$ in point $F$. Point $G$ lies on line segment $AD$ such that $EG$ and $BC$ are parallel. Prove that $|AG| =|BF|$.
[asy]
unitsize (1.5 cm);
real angleindegrees(pair A, pair B, pair C) {
real a, b, c;
a = abs(B - C);
b = abs(C - A);
c = abs(A - B);
return(aCos((a^2 + c^2 - b^2)/(2*a*c)));
};
pair A, B, C, D, E, F, G;
B = (0,0);
A = 2*dir(190);
D = 2*dir(310);
C = 1.5*dir(310 - 180);
E = extension(B, incenter(A,B,C), A, rotate(angleindegrees(A,C,B),A)*(B));
F = extension(B,E,A,C);
G = extension(E, E + D - B, A, D);
filldraw(anglemark(A,C,B,8),gray(0.8));
filldraw(anglemark(B,A,E,8),gray(0.8));
draw(C--A--B);
draw(E--A--D);
draw(interp(C,D,-0.1)--interp(C,D,1.1));
draw(interp(E,B,-0.2)--interp(E,B,1.2));
draw(E--G);
dot("$A$", A, SW);
dot("$B$", B, NE);
dot("$C$", C, NE);
dot("$D$", D, NE);
dot("$E$", E, N);
dot("$F$", F, N);
dot("$G$", G, SW);
[/asy]
In a triangle $ABC$, let $O$ be the circumcenter, $H$ the orthocenter, and $M$ the midpoint of the segment $AH$. The perpendicular at $M$ onto $OM$ intersects lines $AB$ and $AC$ at $P$ and $Q$, respectively. Prove that $MP = MQ$.
Let $ABCD$ be a rectangle with $P$ lies on the segment $AC$. Denote $Q$ as a point on minor arc $PB$ of $(PAB)$ such that $QB = QC$. Denote $R$ as a point on minor arc $PD$ of $(PAD)$ such that $RC = RD$. The lines $CB$, $CD$ meet $(CQR)$ again at $M, N$ respectively. Prove that $BM = DN$.
by Tran Quang Hung
Given a triangle $ABC$ with obtuse $\angle A$ and attitude $AH$ with $H \in BC$. Let $E,F$ on $CA$, $AB$ satisfying $\angle BEH = \angle C$ and $\angle CFH = \angle B$. Let $BE$ cut $CF$ at $D$. Prove that $DE = DF$.
Triangle $ABC$ is equilateral. The point $D$ lies on the extension of $AB$ beyond $B$, the point $E$ lies on the extension of $CB$ beyond $B$, and $|CD| = |DE|$. Prove that $|AD| = |BE|$.
[img]https://1.bp.blogspot.com/-QnAXFw3ijn0/XzR0YjqBQ3I/AAAAAAAAMU0/0TvhMQtBNjolYHtgXsQo2OPGJzEYSfCwACLcBGAsYHQ/s0/2015%2BMohr%2Bp3.png[/img]
In triangle $ABC$ it is known that $\angle ACB = 2\angle ABC$. Furthermore $P$ is an interior point of the triangle $ABC$ such that $AP = AC$ and $PB = PC$. Prove that $\angle BAC = 3 \angle BAP$.
Given a convex pentagon $ABCDE$ in which $\angle ABC = \angle AED = 90^o$, $\angle BAC= \angle DAE$. Let $K$ be the midpoint of the side $CD$, and $P$ the intersection point of lines $AD$ and $BK$, $Q$ be the intersection point of lines $AC$ and $EK$. Prove that $BQ = PE$.
Let $\Gamma$ be the circumscribed circle of a triangle $ABC$ and let $D$ be a point at line segment $BC$. The circle passing through $B$ and $D$ tangent to $\Gamma$ and the circle passing through $C $and $D$ tangent to $\Gamma$ intersect at a point $E \ne D$. The line $DE$ intersects $\Gamma$ at two points $X$ and $Y$ . Prove that $|EX| = |EY|$.
Point $P$ is chosen inside triangle $ABC$ so that $\angle APC+\angle ABC=180^o$ and $BC=AP.$ On the side $AB$, a point $K$ is chosen such that $AK = KB + PC$. Prove that $CK \perp AB$.
Let $\Gamma$ is a circle with diameter $AB$. Let $\ell$ be the tangent of $\Gamma$ at $A$, and $m$ be the tangent of $\Gamma$ through $B$. Let $C$ be a point on $\ell$, $C \ne A$, and let $q_1$ and $q_2$ be two lines that passes through $C$. If $q_i$ cuts $\Gamma$ at $D_i$ and $E_i$ ($D_i$ is located between $C$ and $E_i$) for $i = 1, 2$. The lines $AD_1, AD_2, AE_1, AE_2$ intersects $m$ at $M_1, M_2, N_1, N_2$ respectively. Prove that $M_1M_2 = N_1N_2$.
About the pentagon $ABCDE$ we know that $AB = BC = CD = DE$, $\angle C = \angle D =108^o$, $\angle B = 96^o$. Find the value in degrees of $\angle E$.