Found problems: 509
In the triangle $ABC$ the angle bisector $AD$ is drawn, $E$ is the point of tangency of the inscribed circle to the side $BC$, $I$ is the center of the inscribed circle $\Delta ABC$. The point ${{A} _ {1}}$ on the circumscribed circle $\Delta ABC$ is such that $A {{A} _ {1}} || BC$. Denote by $T$ - the second point of intersection of the line $E {{A} _ {1}}$ and the circumscribed circle $\Delta AED$. Prove that $IT = IA$.
Let $ABCD$ a convex quadrilateral (this means that all interior angles are smaller than $180^o$), such that there exist a point $M$ on line segment $AB$ and a point $N$ on line segment $BC$ having the property that $AN$ cuts the quadrilateral in two parts of equal area, and such that the same property holds for $CM$.
Prove that $MN$ cuts the diagonal $BD$ in two segments of equal length.
Determine the ratio of the bases (parallel sides) of the trapezoid for which there exists a line with $6$ points of intersection with the diagonals, lateral sides and extended bases cut $5$ equal segments?
( E . G . Gotman)
The pentagon $ABCDE$ has $AB = BC, CD = DE, \angle ABC = 120^o, \angle CDE = 60^o$ and $BD = 2$. Calculate the area of the pentagon.
In the cyclic quadrilateral $ABCD$ with $AB = AD$, points $M$ and $N$ lie on the sides $CD$ and $BC$ respectively so that $MN = BN + DM$. Lines $AM$ and $AN$ meet the circumcircle of $ABCD$ again at points $P$ and $Q$ respectively. Prove that the orthocenter of the triangle $APQ$ lies on the segment $MN$.
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 $BP = PQ$.
In triangle $\vartriangle ABC$, $D$ is the midpoint of $BC$. Points $E$ and $F$ are chosen on side $AC$ so that $AF = F E = EC$. Let $AD$ intersect $BE$ and $BF$ and $G$ and $H$, respectively. Find the ratio of the areas of $\vartriangle BGH$ and $\vartriangle ABC$.
A triangle $\vartriangle ABC$ has $\angle A = 90^o$, and a point $D$ is chosen on $AC$. Point $F$ is the foot of altitude from $A$ to $BC$. Suppose that $BD = DC = CF = 2$. Compute $AC$.
In triangle $ABC$, the point $D$ lies on segment $AB$ such that $CD$ is the angle bisector of angle $\angle C$. The perpendicular bisector of segment $CD$ intersects the line $AB$ in $E$. Suppose that $|BE| = 4$ and $|AB| = 5$.
(a) Prove that $\angle BAC = \angle BCE$.
(b) Prove that $2|AD| = |ED|$.
[asy]
unitsize(1 cm);
pair A, B, C, D, E;
A = (0,0);
B = (2,0);
C = (1.8,1.8);
D = extension(C, incenter(A,B,C), A, B);
E = extension((C + D)/2, (C + D)/2 + rotate(90)*(C - D), A, B);
draw((E + (0.5,0))--A--C--B);
draw(C--D);
draw(interp((C + D)/2,E,-0.3)--interp((C + D)/2,E,1.2));
dot("$A$", A, SW);
dot("$B$", B, S);
dot("$C$", C, N);
dot("$D$", D, S);
dot("$E$", E, S);
[/asy]
Let $ABCDEF$ be a hexagon inscribed in a circle in which $AB = BC, CD = DE$ and $EF = FA$. Prove that the lines $AD, BE$ and $CF$ intersect at one point.
On the sides $AB$ and $BC$ of the triangle $ABC$, the points $K$ and $M$ are chosen so that $KM \parallel AC$. The segments $AM$ and $KC$ intersect at the point $O$. It is known that $AK =AO$ and $KM =MC$. Prove that $AM=KB$.
(i) $ABC$ is a triangle with a right angle at $A$, and $P$ is a point on the hypotenuse $BC$.
The line $AP$ produced beyond $P$ meets the line through $B$ which is perpendicular to $BC$ at $U$.
Prove that $BU = BA$ if, and only if, $CP = CA$.
(ii) $A$ is a point on the semicircle $CB$, and points $X$ and $Y$ are on the line segment $BC$.
The line $AX$, produced beyond $X$, meets the line through $B$ which is perpendicular to $BC$ at $U$.
Also the line $AY$, produced beyond $Y$, meets the line through $C$ which is perpendicular to $BC$ at $V$.
Given that $BY = BA$ and $CX = CA$, determine the angle $\angle VAU$.
Non-degenerate quadrilateral $ABCD$ with $AB = AD$ and $BC = CD$ has integer side lengths, and $\angle ABC = \angle BCD = \angle CDA$. If $AB = 3$ and $B \ne D$, how many possible lengths are there for $BC$?
In quadrilateral $ABCD$ sides $BC$ and $AD$ are parallel. In each of the four vertices we draw an angular bisector. The angular bisectors of angles $A$ and $B$ intersect in point $P$, those of angles $B$ and $C$ intersect in point $Q$, those of angles $C$ and $D$ intersect in point $R$, and those of angles $D$ and $A$ intersect in point S. Suppose that $PS$ is parallel to $QR$. Prove that $|AB| =|CD|$.
[asy]
unitsize(1.2 cm);
pair A, B, C, D, P, Q, R, S;
A = (0,0);
D = (3,0);
B = (0.8,1.5);
C = (3.2,1.5);
S = extension(A, incenter(A,B,D), D, incenter(A,C,D));
Q = extension(B, incenter(A,B,C), C, C + incenter(A,B,D) - A);
P = extension(A, S, B, Q);
R = extension(D, S, C, Q);
draw(A--D--C--B--cycle);
draw(B--Q--C);
draw(A--S--D);
dot("$A$", A, SW);
dot("$B$", B, NW);
dot("$C$", C, NE);
dot("$D$", D, SE);
dot("$P$", P, dir(90));
dot("$Q$", Q, dir(270));
dot("$R$", R, dir(90));
dot("$S$", S, dir(90));
[/asy]
Attention: the figure is not drawn to scale.
Circles $S_1$ and $S_2$ with centers $O_1$ and $O_2$ respectively intersect at $A$ and $B$. The circle passing through $O_1$, $O_2$, and $A$ intersects $S_1$, $S_2$ and line $AB$ again at $D$, $E$, and $C$, respectively. Show that $CD = CB = CE$.
In triangle $ABC$, points $K ,P$ are chosen on the side $AB$ so that $AK = BL$, and points $M,N$ are chosen on the side $BC$ so that $CN = BM$. Prove that $KN + LM \ge AC$.
(I. Bogdanov)
$ABCDEF$ is a convex hexagon with all its sides equal. Also $\angle A + \angle C + \angle E = \angle B + \angle D + \angle F$. Show that $\angle A = \angle D$, $\angle B = \angle E$ and $\angle C = \angle F$.
Let $AD$ be the bisector of triangle $ABC$. Circle $\omega$ passes through the vertex $A$ and touches the side $BC$ at point $D$. This circle intersects the sides $AC$ and $AB$ for the second time at points $M$ and $N$ respectively. Lines $BM$ and $CN$ intersect the circle for the second time $\omega$ at points $P$ and $Q$, respectively. Lines $AP$ and $AQ$ intersect side $BC$ at points $K$ and $L$, respectively. Prove that $KL=\frac12 BC$
On side $AB$ of triangle $ABC$ is marked point $K$ such that $AB = CK$. Points $N$ and $M$ are the midpoints of $AK$ and $BC$, respectively. The segments $NM$ and $CK$ intersect in point $P$. Prove that $KN = KP$.
Point $D$ on side $BC$ of acute triangle ABC is such that $AB=AD$. The circumcircle of triangle $ABD$ intersects side $AC$ at points $A$ and $K$. Line $DK$ intersects the perpendicular drawn from $B$ on $AC$, at the point $L$. Prove that $CL= BC$
Consider a quadrilateral $ABCD$ and points $K, L, M, N$ on sides $AB, BC, CD$ and $AD$, respectively, such that $KB = BL = a, MD = DN = b$ and $KL \nparallel MN$. Find the set of all the intersection points of $KL$ with $MN$ as $a$ and $b$ vary.
In the quadrilateral $ABCD$ the condition $AD = AB + CD$ is fulfilled. The bisectors of the angles $BAD$ and $ADC$ intersect at the point $P $, as shown in Fig. Prove that $BP = CP$.
[img]https://cdn.artofproblemsolving.com/attachments/3/1/67268635aaef9c6dc3363b00453b327cbc01f3.png[/img]
(Maria Rozhkova)
Point $L$ is marked on the side $AB$ of a triangle $ABC$. The incircle of the triangle $ABC$ meets the segment $CL$ at points $P$ and $Q$ .Is it possible that the equalities $CP = PQ = QL$ hold if $CL$ is
a) the median?
b) the bisector?
c) the altitude?
d) the segment joining vertex $C$ with the point $L$ of tangency of the excircle of the triangie $ABC$ with $AB$ ?
(I. Gorodnin)
In a triangle $ABC$ we have $|AB| = |AC|$ and $\angle BAC = \alpha$. Let $P \ne B$ be a point on $AB$ and $Q$ a point on the altitude drawn from $A$ such that $|PQ| = |QC|$. Find $ \angle QPC$.
Let $ABCD$ be a rectangle and the circle of center $D$ and radius $DA$, which cuts the extension of the side $AD$ at point $P$. Line $PC$ cuts the circle at point $Q$ and the extension of the side $AB$ at point $R$. Show that $QB = BR$.