Found problems: 509
In trapezoid $ABCD$, the bisectors of angles $A$ and $D$ intersect at point $E$ lying on the side of $BC$. These bisectors divide the trapezoid into three triangles into which the circles are inscribed. One of these circles touches the base $AB$ at the point $K$, and two others touch the bisector $DE$ at points $M$ and $N$. Prove that $BK = MN$.
In a triangle $ABC$, let $K$ be a point on the median $BM$ such that $CM = CK$. It turned out that $\angle CBM = 2\angle ABM$. Show that $BC = KM$.
The points $D, E, F$ lie respectively on the sides $BC$, $CA$, $AB$ of the triangle ABC such that $F B = BD$, $DC = CE$, and the lines $EF$ and $BC$ are parallel. Tangent to the circumscribed circle of triangle $DEF$ at point $F$ intersects line $AD$ at point $P$. Perpendicular bisector of segment $EF$ intersects the segment $AC$ at $Q$. Prove that the lines $P Q$ and $BC$ are parallel.
The altitueds $A {{A} _ {1}} $, $B {{B} _ {1}}$ and $C {C} _ 1$ are drawn in the acute triangle $ABC$. . The perpendicular $AK$ is drawn from the vertex $A$ on the line ${{A} _ {1}} {{B} _ {1}}$, and the perpendicular $BL$ is drawn from the vertex $B$ on the line ${{C} _ {1}} {{B} _ {1}}$. Prove that ${{A} _ {1}} K = {{B} _ {1}} L$.
(Maria Rozhkova)
Let $ABC$ be an acute-angled triangle, $M$ the midpoint of the side $AC$ and $F$ the foot on $AB$ of the altitude through the vertex $C$. Prove that $AM = AF$ holds if and only if $\angle BAC = 60^o$.
(Karl Czakler)
In the trapezoid $ABCD, AB // CD$ and the diagonals intersect at $O$. The points $P, Q$ are on $AD, BC$ respectively such that $\angle AP B = \angle CP D$ and $\angle AQB = \angle CQD$. Show that $OP = OQ$.
The diagonals of the inscribed quadrilateral $ABCD$ meet at the point $M$, $\angle AMB = 60^o$. Equilateral triangles $ADK$ and $BCL$ are built outward on sides $AD$ and $BC$. Line $KL$ meets the circle circumscribed ariound $ABCD$ at points $P$ and $Q$. Prove that $PK = LQ$.
Let $\Gamma$ be a semicircle with center $O$ and diameter $AB$. $D$ is the midpoint of arc $AB$. On the ray $OD$, we take $E$ such that $OE = BD$. $BE$ intersects the semicircle at $F$ and $ P$ is the point on $AB$ such that $FP$ is perpendicular to $AB$. Prove that $BP=\frac13 AB$.
Let $ABC$ be an acute, non isosceles triangle and $(O)$ be its circumcircle (with center $O$). Denote by $G$ the centroid of the triangle $ABC$, by $H$ the foot of the altitude from $A$ onto the side $BC$ and by $I$ the midpoint of $AH$. The line $IG$ intersects $BC$ at $K$.
1. Prove that $CK = BH$.
2. The ray $(GH$ intersects $(O)$ at L. Denote by $T$ the circumcenter of the triangle $BHL$. Prove that $AO$ and $BT$ intersect on the circle $(O)$.
Natural $p$ and $q$ are relatively prime. The $[0,1]$ is divided onto $(p+q)$ equal segments. Prove that every segment except two marginal contain exactly one from the $(p+q-2)$ numbers $$\{1/p, 2/p, ... , (p-1)/p, 1/q, 2/q, ... , (q-1)/q\}$$
Circles $S_1$ and $S_2$ with centers $O_1$ and $O_2$ respectively intersect at $A$ and $B$. Ray $O_1B$ meets $S_2$ again at $F$, and ray $O_2B$ meets $ S_1$ again at $E$. The line through $B$ parallel to $ EF$ intersects $S_1$ and $S_2$ again at $M$ and $N$, respectively. Prove that $MN = AE +AF$.
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)
Prove that in a triangle one of the sides is twice as large as the other if and only if a median and an angle bisector of this triangle are perpendicular
Let $ABCD$ be a convex quadrilateral with the property that $MA \cdot MC + MA \cdot CD = MB \cdot MD$, where $M$ is the intersection of the diagonals $AC$ and $BD$. The angle bisector of $\angle ACD$ is drawn intersecting ray $\overrightarrow{BA}$ at $K$. Prove that $BC = DK$ if and only if $AB \parallel CD$.
It is known that $ABCD$ is a parallelogram. The point $E$ is taken so that $BCED$ is a cyclic quadrilateral. Let $\ell$ be a line that passes through $A$, intersects the segment $DC$ at point $F$ and intersects the extension of the line $BC$ at $G$. Given $EF = EG = EC$. Prove that $\ell$ is the bisector of the angle $\angle BAD$.
We consider an acute triangle $ABC$. The altitude from $A$ is $AD$, the altitude from $D$ in triangle $ABD$ is $DE,$ and the altitude from $D$ in triangle $ACD$ is $DF$.
a) Prove that the triangles $ABC$ and $AF E$ are similar.
b) Prove that the segment $EF$ and the corresponding segments constructed from the vertices $B$ and $C$ all have the same length.
Let $t$ be a line passing through the vertex $A$ of the equilateral $ABC$, parallel to the side $BC$. On the side $AC$ arbitrarily mark the point $D$. Bisector of the angle $ABD$ intersects the line $t$at the point $E$. Prove that $BD=CD+AE$.
On the sides $AC$ and $AB$ of the triangle $ABC$, the points $D$ and $E$ were chosen such that $\angle ABD =\angle CBD$ and $3 \angle ACE = 2\angle BCE$. Let $H$ be the point of intersection of $BD$ and $CE$, and $CD = DE = CH$. Find the angles of triangle $ABC$.
In any $\vartriangle ABC, \ell$ is any line through $C$ and points $P, Q$. If $BP, AQ$ are perpendicular to the line $\ell$ and $M$ is the midpoint of the line segment $AB$, then prove that $MP = MQ$
Let $ABC$ be a triangle and $AD$ be the angle bisector of $<BAC$, with $D$ on $BC$. Let $E$ be a point on segment $BC$ such that $BD = EC$. Through $E$ draw $l$ a parallel line to $AD$ and let $P$ be a point in $l$ inside the triangle. Let $G$ be the point where $BP$ intersects $AC$ and $F$ be the point where $CP$ intersects $AB$. Show $BF = CG$.
The quadrilateral $ABCD$ is inscribed in the circle and the lengths of the sides $BC$ and $DC$ are equal, and the length of the side $AB$ is equal to the length of the diagonal $AC$. Let the point $P$ be the midpoint of the arc $CD$, which does not contain point $A$, and $Q$ is the point of intersection of diagonals $AC$ and $BD$. Prove that the lines $PQ$ and $AB$ are perpendicular.
The angle bisector of $A$ in triangle $ABC$ intersects $BC$ in the point $D$. The point $E$ lies on the side $AC$, and the lines $AD$ and $BE$ intersect in the point $F$. Furthermore, $\frac{|AF|}{|F D|}= 3$ and $\frac{|BF|}{|F E|}=\frac{5}{3}$. Prove that $|AB| = |AC|$.
[img]https://1.bp.blogspot.com/-evofDCeJWPY/XzT9dmxXzVI/AAAAAAAAMVY/ZN87X3Cg8iMiULwvMhgFrXbdd_f1f-JWwCLcBGAsYHQ/s0/2013%2BMohr%2Bp5.png[/img]
In an acute-angled triangle $ABC, E$ and $F$ are the feet of the altitudes from $B$ and $C$, and $G$ and $H$ are the projections of $B$ and $C$ on $EF$, respectively. Prove that $HE = FG$.
On sides $AB$ and $CB$ of $\vartriangle ABC$ there are drawn equal segments, $AD$ and $CE$, respectively, of arbitrary length (but shorter than min($AB,BC$)). Find the locus of midpoints of all possible segments $DE$.
Given points $D$ and $E$ on the legs $[CA]$ and $[CB]$, respectively, of the isosceles right triangle. $|CD| = |CE|$. The extensions of the perpendiculars from $D$ and $C$ to the line $AE$ cross the hypotenuse $AB$ in the points $K$ and $L$. Prove that $|KL| = |LB|$