The trapezium ${ABCD}$, where ${AB}$ and ${CD}$ are parallel and ${AD < CD}$, is inscribed in the circle ${c}$. Let ${DP}$ be a chord of the circle, parallel to ${AC}$. Assume that the tangent to ${c}$ at ${D}$ meets the line ${AB}$ at ${E}$ and that ${PB}$ and ${DC}$ meet at ${Q}$. Show that ${EQ = AC}$.

Isosceles triangles with a fixed angle $\alpha$ at the vertex opposite to the base are being inscribed into a rectangle $ABCD$ so that this vertex lies on the side $BC$ and the vertices of the base lie on the sides $AB$ and $CD$. Prove that the midpoints of the bases of all such triangles coincide. (Igor Zhizhilkin)

Given a point $A$ that lies on circle $c(o,R)$ (with center $O$ and radius $R$). Let $(e)$ be the tangent of the circle $c$ at point $A$ and a line $(d)$ that passes through point $O$ and intersects $(e)$ at point $M$ and the circle at points $B,C$ (let $B$ lie between $O$ and $A$). If $AM = R\sqrt3$ , prove that a) triangle $AMC$ is isosceles. b) circumcenter of triangle $AMC$ lies on circle $c$ .

In the triangle $ABC$, the bisector of angle $A$ is equal to the half-sum of the height and median drawn from vertex $A$. Prove that if $\angle A$ is obtuse, then $AB = AC$.

Pinocchio claims that he can divide an isoceles triangle into three triangles, any two of which can be put together to form a new isosceles triangle. Is Pinocchio lying? (A Shapovalov)

In a trapezoid, the length of one of the diagonals is equal to the sum of the lengths of the bases, and the angle between the diagonals is $60$ degrees. Prove that this trapezoid is isosceles.

In the rectangle $ABCD$ with $|AB|<2 |AD|$, let $E$ be the midpoint of $AB$ and $F$ a point on the chord $CE$ such that $\angle CFD = 90^o$. Prove that $FAD$ is an isosceles triangle.

Let $ABC$ be an isosceles triangle with $AC = BC$. On the arc $CA$ of its circumcircle, which does not contain $ B$, there is a point $ P$. The projection of $C$ on the line $AP$ is denoted by $E$, the projection of $C$ on the line $BP$ is denoted by $F$. Prove that the lines $AE$ and $BF$ have equal lengths. (W. Janous, WRG Ursulincn, Innsbruck)

Consider an isosceles triangle $KL_1L_2$ with $|KL_1|=|KL_2|$ and let $KA, L_1B_1,L_2B_2$ be its angle bisectors. Prove that $\cos \angle B_1AB_2 < \frac35$

Let $ABC$ a right isosceles triangle with hypotenuse $AB=\sqrt2$ . Determine the positions of the points $X,Y,Z$ on the sides $BC,CA,AB$ respectively so that the triangle $XYZ$ is isosceles, right, and with minimum area.

Let $ABCD$ be a cyclic quadrangle, let the diagonals $AC$ and $BD$ cross at $O$, and let $I$ and $J$ be the incentres of the triangles $ABC$ and $ABD$, respectively. The line $IJ$ crosses the segments $OA$ and $OB$ at $M$ and $N$, respectively. Prove that the triangle $OMN$ is isosceles.

Let $ABC$ be an acute triangle with $AB\le AC$ and let $c(O,R)$ be it's circumscribed circle (with center $O$ and radius $R$). The perpendicular from vertex $A$ on the tangent of the circle passing through point $C$, intersect it at point $D$. a) If the triangle $ABC$ is isosceles with $AB=AC$, prove that $CD=BC/2$. b) If $CD=BC/2$, prove that the triangle $ABC$ is isosceles.

At least how many regular triangles are needed to cover the lines of the following diagram? [img]https://cdn.artofproblemsolving.com/attachments/e/3/4de2ed2c7cc9421d7d060f0bc537ccaa3838fc.png[/img] (Only the perimeter of the triangles is involved in the covering, and the entire perimeter need not be incident on the diagram.)

In a right-angled and isosceles triangle, the two catheti are both length $1$. Find the length of the shortest line segment dividing the triangle into two figures with the same area, and specify the location of this line segment

On a isosceles triangle $\triangle ABC$ with $AB=BC$ let $K,M$ be the midpoints of $AB,AC$ respectivily. Let $(CKB)$ intersect $BM$ at $N \ne M$, the line through $N$ parallel to $AC$ intersects $(ABC)$ at $A_1,C_1$. Show that $\triangle A_1BC_1$ is equilateral.

In triangle $ABC$ the bisector of the angle at $B$ meets $AC$ at $D$ and the bisector of the angle at $C$ meets $AB$ at $E$. These bisectors intersect at $O$ and the lengths of $OD$ and $OE$ are equal. Prove that either $\angle BAC = 60^o$ or triangle $ABC$ is isosceles.

Let $ABC$ be a non isosceles triangle inscribed in a circle $(O)$ and $BE, CF$ are two angle bisectors intersect at $I$ with $E$ belongs to segment $AC$ and $F$ belongs to segment $AB$. Suppose that $BE, CF$ intersect $(O)$ at $M,N$ respectively. The line $d_1$ passes through $M$ and perpendicular to $BM$ intersects $(O)$ at the second point $P,$ the line $d_2$ passes through $N$ and perpendicular to $CN$ intersect $(O)$ at the second point $Q$. Denote $H, K$ are two midpoints of $MP$ and $NQ$ respectively. 1. Prove that triangles $IEF$ and $OKH$ are similar. 2. Suppose that S is the intersection of two lines $d_1$ and $d_2$. Prove that $SO$ is perpendicular to $EF$.

Let $ABC$ be an non- isosceles triangle, $H_a$, $H_b$, and $H_c$ be the feet of the altitudes drawn from the vertices $A, B$, and $C$, respectively, and $M_a$, $M_b$, and $M_c$ be the midpoints of the sides $BC$, $CA$, and $AB$, respectively. The circumscribed circles of triangles $AH_bH_c$ and $AM_bM_c$ intersect for second time at point $A'$. The circumscribed circles of triangles $BH_cH_a$ and $BM_cM_a$ intersect for second time at point $B'$. The circumscribed circles of triangles $CH_aH_b$ and $CM_aM_b$ intersect for second time at point $C'$. Prove that points $A', B'$ and $C'$ lie on the same line.

In an acute-angled isosceles triangle $ABC$, altitudes $CC_1$ and $BB_1$ intersect the line passing through the vertex $A$ and parallel to the line $BC$, at points $P$ and $Q$. Let $A_0$ be the midpoint of side $BC$, and $AA_1$ the altitude. Lines $A_0C_1$ and $A_0B_1$ intersect line $PQ$ at points $K$ and $L$. Prove that the circles circumscribed around triangles $PQA_1, KLA_0, A_1B_1C_1$ and a circle with a diameter $AA_1$ intersect at one point.

Let $d$ be the tangent at $B$ to the circumcircle of the acute scalene triangle $ABC$. Let $K$ be the orthogonal projection of the orthocenter, $H$, of triangle $ABC$ to the line $d$ and $L$ the midpoint of the side $AC$. Prove that the triangle $BKL$ is isosceles.

On the sides of triangle $ABC$, isosceles right-angled triangles $AUB, CVB$, and $AWC$ are placed. These three triangles have their right angles at vertices $U, V$ , and $W$, respectively. Triangle $AUB$ lies completely inside triangle $ABC$ and triangles $CVB$ and $AWC$ lie completely outside $ABC$. See the figure. Prove that quadrilateral $UVCW$ is a parallelogram. [asy] import markers; unitsize(1.5 cm); pair A, B, C, U, V, W; A = (0,0); B = (2,0); C = (1.7,2.5); U = (B + rotate(90,A)*(B))/2; V = (B + rotate(90,C)*(B))/2; W = (C + rotate(90,A)*(C))/2; draw(A--B--C--cycle); draw(A--W, StickIntervalMarker(1,1,size=2mm)); draw(C--W, StickIntervalMarker(1,1,size=2mm)); draw(B--V, StickIntervalMarker(1,2,size=2mm)); draw(C--V, StickIntervalMarker(1,2,size=2mm)); draw(A--U, StickIntervalMarker(1,3,size=2mm)); draw(B--U, StickIntervalMarker(1,3,size=2mm)); draw(rightanglemark(A,U,B,5)); draw(rightanglemark(B,V,C,5)); draw(rightanglemark(A,W,C,5)); dot("$A$", A, S); dot("$B$", B, S); dot("$C$", C, N); dot("$U$", U, NE); dot("$V$", V, NE); dot("$W$", W, NW); [/asy]

Let $ABCD$ be an isosceles trapezoid with bases $AB$ and $CD$. The incircle of the triangle $BCD$ touches $CD$ at $E$. Point $F$ is chosen on the bisector of the angle $DAC$ such that the lines $EF$ and $CD$ are perpendicular. The circumcircle of the triangle $ACF$ intersects the line $CD$ again at $G$. Prove that the triangle $AFG$ is isosceles.

The centre of circle $1$ lies on circle $2$. $A$ and $B$ are the intersection points of the circles. The tangent line to circle $2$ at point $B$ intersects circle $1$ at point $C$. Prove that $AB = BC$. (V. Prasovov, Moscow)

Let $ABC$ be an isosceles triangle with $AC = BC$, and let $M$ be the midpoint of $AB$. Let $P$ be a point inside the triangle such that $\angle PAB =\angle PBC$. Prove that $\angle APM + \angle BPC = 180^o$.

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]