Found problems: 509
Given a quadrilateral $ABCD$ with $AB = BD = DC$ and $AC = BC$. On $BC$ lies point $E$ such that $AE = AB$. Prove that $ED = EB$.
Five points $A, B, C, D$, and $E$ in three-dimensional Euclidean space have the property that $AB = BC = CD = DE = EA = 1$ and $\angle ABC = \angle BCD =\angle CDE = \angle DEA = 90^o$ . Find all possible $\cos(\angle EAB)$.
Let $ABC$ be a triangle with $\angle CAB = 2 \angle ABC$. Assume that a point $D$ is inside the triangle $ABC$ exists such that $AD = BD$ and $CD = AC$. Show that $\angle ACB = 3 \angle DCB$.
The inscribed circle of the triangle $ABC$ touches the sides $BC, CA, AB$ at points $A_1, B_1, C_1$ respectively. The points $P_b, Q_b, R_b$ are the points of the segments $BC_1, C_1A_1, A_1B$, respectively, such that $BP_bQ_bR_b$ is parallelogram. In the same way, the points $P_c, Q_c, R_c$ are the points of the sections $CB_1, B_1A_1, A_1C$, respectively such that $CP_cQ_cR_c$ is a parallelogram. The intersection of the lines $P_bR_b$ and $P_cR_c$ is $T$. Show that $TQ_b = TQ_c$.
$ABC$ is a triangle, $H$ its orthocenter, $I$ its incenter, $O$ its circumcenter and $\omega$ its circumcircle. Line $CI$ intersects circle $\omega$ at point $D$ different from $C$. Assume that $AB = ID$ and $AH = OH$. Find the angles of triangle $ABC$.
Does there exist a nonisosceles triangle such that the altitude from one vertex, the bisectrix from the second one and the median from the third one are equal?
The bisector of angle $B$ and the bisector of external angle $D$ of rectangle $ABCD$ intersect side $AD$ and line $AB$ at points $M$ and $K$, respectively. Prove that the segment $MK$ is equal and perpendicular to the diagonal of the rectangle.
In the figure below the triangles $BCD, CAE$ and $ABF$ are equilateral, and the triangle $ABC$ is right-angled with $\angle A = 90^o$. Prove that $|AD| = |EF|$.
[img]https://1.bp.blogspot.com/-QMMhRdej1x8/XzP18QbsXOI/AAAAAAAAMUI/n53OsE8rwZcjB_zpKUXWXq6bg3o8GUfSwCLcBGAsYHQ/s0/2019%2Bmohr%2Bp5.png[/img]
Let $ABCD$ be a parallelogram with an acute angle at $A$. Let $G$ be a point on the line $AB$, distinct from $B$, such that $|CG| = |CB|$. Let $H$ be a point on the line $BC$, distinct from $B$, such that $|AB| =|AH|$. Prove that triangle $DGH$ is isosceles.
[asy]
unitsize(1.5 cm);
pair A, B, C, D, G, H;
A = (0,0);
B = (2,0);
D = (0.5,1.5);
C = B + D - A;
G = reflect(A,B)*(C) + C - B;
H = reflect(B,C)*(H) + A - B;
draw(H--A--D--C--G);
draw(interp(A,G,-0.1)--interp(A,G,1.1));
draw(interp(C,H,-0.1)--interp(C,H,1.1));
draw(D--G--H--cycle, dashed);
dot("$A$", A, SW);
dot("$B$", B, SE);
dot("$C$", C, E);
dot("$D$", D, NW);
dot("$G$", G, NE);
dot("$H$", H, SE);
[/asy]
A quadrilateral $ABCD$ is inscribed in a circle and its two diagonals $AC,BD$ meet at $G$. Let $M$ be the center of $CD, E,F$ be the points on $BC, AD$ respectively such that $ME \parallel AC$ and $MF \parallel BD$. Point $H$ is the projection of $G$ onto $CD$. The circumcircle of $MEF$ meets $CD$ at $N$ distinct from $M$. Prove that $MN = MH$
Tran Quang Hung, Nguyen Le Phuoc, Thanh Xuan, Hanoi
On the legs $AC$ and $BC$ of an isosceles right-angled triangle with a right angle $C$, points $D$ and $E$ are taken, respectively, so that $CD = CE$. Perpendiculars on line $AE$ from points $C$ and $D$ intersect segment $AB$ at points $P$ and $Q$, respectively. Prove that $BP = PQ$.
The vertex $F$ of the parallelogram $ACEF$ lies on the side $BC$ of parallelogram $ABCD$. It is known that $AC = AD$ and $AE = 2CD$. Prove that $\angle CDE = \angle BEF$.
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 $|EX| = |EY|$.
Let $M$ be a point on the diameter $AB$ of a semicircle $\Gamma$. The perpendicular at $M$ meets the semicircle $\Gamma$ at $P$. A circle inside $\Gamma$. touches $\Gamma$. and is tangent to $PM$ at $Q$ and $AM$ at $R$. Prove that $P B = RB$.
On the diagonals $AC$ and $CE$ of a regular hexagon $ABCDEF$ with side $1$ we mark points$ M$ and $N$ such that $AM = CN = a$. Find $a$ if the points $B, M, N$ lie on the same line.
Let $ABC$ be an acute triangle and $A_1, B_1$ and $C_1$, points on the sides $BC, CA$ and $AB$, respectively, such that $CB_1 = A_1B_1$ and $BC_1 = A_1C_1$. Let $D$ be the symmetric of $A_1$ with respect to $B_1C_1, O$ and $O_1$ are the circumcenters of triangles $ABC$ and $A_1B_1C_1$, respectively. If $A \ne D, O \ne O_1$ and $AD$ is perpendicular to $OO_1$, prove that $AB = AC$.
Given a $ \vartriangle ABC $ with $ AB> AC $ and $ \angle BAC = 60^o$. Denote the circumcenter and orthocenter as $ O $ and $ H $ respectively. We also have that $ OH $ intersects $ AB $ in $ P $ and $ AC $ in $ Q $. Prove that $ PO = HQ $.
In the acute-angled triangle $ABC$, the altitudes $BD$ and $CE$ are drawn. Let $F$ and $G$ be the points of the line $ED$ such that $BF$ and $CG$ are perpendicular to $ED$. Prove that $EF = DG$.
Let $ABC$ be a triangle with $ \angle ACB = 90^o + \frac12 \angle ABC$ . The point $M$ is the midpoint of the side $BC$ . A circle with center at vertex $A$ intersects the line $BC$ at points $M$ and $D$. Prove that $MD = AB$.
Above the segments $AB$ and $BC$ we drew a semicircle at each. $F_1$ bisects $AB$ and $F_2$ bisects $BC$. Above the segments $AF_2$ and $F_1C$ we also drew a semicircle at each. Segments $P Q$ and $RS$ touch the corresponding semicircles as shown in the figure. Prove that $P Q \parallel RS$ and $|P Q| = 2 \cdot |RS|$.
[img]https://cdn.artofproblemsolving.com/attachments/8/2/570e923b91e9e630e3880a014cc6df4dc33aa2.png[/img]
An arbitrary point $D$ was marked on the median $BM$ of the triangle $ABC$. It is known that the point $DE\parallel AB$ and $CE \parallel BM$. Prove that $BE = AD$
Let $\vartriangle ABC$ be an equilateral triangle, and let $M$ and $N$ be points on $AB$ and $AC$, respectively, so that $AN = BM$ and $3MB = AB$. Lines $CM$ and $BN$ intersect at $O$. Find $\angle AOB$.
Consider in the interior of an equilateral triangle $ABC$ points $D, E$ and $F$ such that$ D$ belongs to segment $BE$, $E$ belongs to segment $CF$ and$ F$ to segment $AD$. If $AD=BE = CF$ then $DEF$ is equilateral.
An isosceles triangle $ABC$ with base $AC$ is given. On the rays $CA$, $AB$ and $BC$, the points $D, E$ and $F$ were marked, respectively, in such a way that $AD = AC$, $BE = BA$ and $CF = CB$. Find the sum of the angles $\angle ADB$, $\angle BEC$ and $\angle CFA$.
Rectangle $ABCD$ with $AB>BC$ has point $P$ inside of it and $Q$ outside of it, such that $PQCD$ is a parallelogram with $PD=AD$. Let $M$ be the midpoint of $CD$. Give that $\angle AMP=\angle BMQ$, prove that $AB=2BC$.