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 $CK \perp AB$.

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]

Let $Z$ and $Y$ be the tangency points of the incircle of the triangle $ABC$ with the sides $AB$ and $CA$, respectively. The parallel line to $Y Z$ through the midpoint $M$ of $BC$, meets $CA$ in $N$. Let $L$ be the point in $CA$ such that $NL = AB$ (and $L$ on the same side of $N$ than $A$). The line $ML$ meets $AB$ in $K$. Prove that $KA = NC$.

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 $ABC$.

Point $K$ is marked on the diagonal $AC$ in rectangle $ABCD$ so that $CK = BC$. On the side $BC$, point $M$ is marked so that $KM = CM$. Prove that $AK + BM = CM$.

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 $\cos(\angle EAB)$.

Inside the equilateral triangle $ABC$, points $P$ and $Q$ are chosen so that the quadrilateral $APQC$ is convex, $AP = PQ = QC$ and $\angle PBQ = 30^o$. Prove that $AQ = BP$.

On the side $AB$ of the non-isosceles triangle $ABC$, let the points $P$ and $Q$ be so that $AC = AP$ and $BC = BQ$. The perpendicular bisector of the segment $PQ$ intersects the angle bisector of the $\angle C$ at the point $R$ (inside the triangle). Prove that $\angle ACB + \angle PRQ = 180^o$.

An arbitrary point $M$ inside an equilateral triangle $ABC$ was connected to vertices. Prove that on each side the triangle can be selected one point at a time so that the distances between them would be equal to $AM, BM, CM$.

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]

Given a semicirle with center $O$ an arbitrary inner point of the diameter divides it into two segments. Let there be semicircles above the two segments as visible in the below figure. The line $\ell$ passing through the point $A$ intersects the semicircles in $4$ points: $B, C, D$ and $E$. Show that the segments $BC$ and $DE$ have the same length. [img]https://cdn.artofproblemsolving.com/attachments/1/4/86a369d54fef7e25a51fea6481c0b5e7dd45ff.png[/img]

Let $ABC$ be a triangle inscribed in the circle $(O)$. Tangents at $B,C$ of the circles $(O)$ meet at $T$ . Let $M,N$ be the points on the rays $BT,CT$ respectively such that $BM = BC = CN$. The line through $M$ and $N$ intersects $CA,AB$ at $E, F$ respectively; $BE$ meets $CT$ at $P, CF$ intersects $BT$ at $Q$. Prove that $AP = AQ$. Trần Quang Hùng

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 $|AS| = |SP|$.

In triangle $ABC$, let $I$ be the incentre, $O$ be the circumcentre, and $H$ be the orthocentre. It is given that $IO = IH$. Show that one of the angles of triangle $ABC$ must be equal to $60$ degrees.

In the right triangle $ABC$ ( $\angle A = 90^o$), $D$ is the intersection of the bisector of the angle $A$ with the side $(BC)$, and $P$ and $Q$ are the projections of the point $D$ on the sides $(AB),(AC)$ respectively . If $BQ \cap DP=\{M\}$, $CP \cap DQ=\{N\}$, $BQ\cap CP=\{H\}$, show that: a) $PM = DN$ b) $MN \parallel BC$ c) $AH \perp BC$.

In a right triangle $ABC$, $K$ is the midpoint of the hypotenuse $AB$ and $M$ such a point on the $BC$ that $| B M | = 2 | MC |$. Prove that $\angle MAB = \angle MKC$.

The measure of the angle $\angle A$ of the acute triangle $ABC$ is $60^o$, and $HI = HB$, where $I$ and $H$ are the incenter and the orthocenter of the triangle $ABC$. Find the measure of the angle $\angle B$.

Let $ABC$ be a triangle. On the extensions of sides $AB$ and $CB$ towards $B$, points $C_1$ and $A_1$ are taken, respectively, so that $AC = A_1C = AC_1$. Prove that circumscribed circles of triangles $ABA_1$ and $CBC_1$ intersect on the bisector of angle $B$.

Let $S$ be the incenter of the triangle $ABC$ and let the line $AS$ intersect the circumcircle of triangle $ABC$ at point $D$ ($D\ne A$). Prove that the segments $BD, CD$ and $SD$ are of equal length.

Given two circles $\omega_1$ and $\omega_2$, with centers $O_1$ and $O_2$, respectively intesrecting at two points $A$ and $B$. Let $X$ and $Y$ be points on $\omega_1$. The lines $XA$ and $YA$ intersect $\omega_2$ again in $Z$ and $W$, respectively, such that $A$ is between $X,Z$ and $A$ is between $Y,W$. Let $M$ be the midpoint of $O_1O_2$, S be the midpoint of $XA$ and $T$ be the midpoint of $WA$. Prove that $MS = MT$ if, and only if, the points $X, Y, Z$ and $W$ are concyclic.

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 $\angle AOB$.

In Fig. , $PA$ and $QB$ are tangents to the circle at $A$ and $B$ respectively. The line $AB$ is extended to meet $PQ$ at $S$. Suppose that $PA = QB$. Prove that $QS = SP$. [img]https://cdn.artofproblemsolving.com/attachments/6/f/f21c0c70b37768f3e80e9ee909ef34c57635d5.png[/img]

Let $ABC$ be an acute triangle such that $\angle BAC = 45^o$. Let $D$ a point on $AB$ such that $CD \perp AB$. Let $P$ be an internal point of the segment $CD$. Prove that $AP\perp BC$ if and only if $|AP| = |BC|$.

The side $BC$ of the triangle $ABC$ is extended beyond $C$ to $D$, such that $CD=BC$. The side $CA$ is extended beyond $A$ to $E$, such that $AE=2CA$. Prove that if $AD=BE$, then the triangle $ABC$ is right.

Let $AB$ be a diameter of a circle $(\omega)$ with centre $O$. From an arbitrary point $M$ on $AB$ such that $MA < MB$ we draw the circles $(\omega_1)$ and $(\omega_2)$ with diameters $AM$ and $BM$ respectively. Let $CD$ be an exterior common tangent of $(\omega_1), (\omega_2)$ such that $C$ belongs to $(\omega_1)$ and $D$ belongs to $(\omega_2)$. The point $E$ is diametrically opposite to $C$ with respect to $(\omega_1)$ and the tangent to $(\omega_1)$ at the point $E$ intersects $(\omega_2)$ at the points $F, G$. If the line of the common chord of the circumcircles of the triangles $CED$ and $CFG$ intersects the circle $(\omega)$ at the points $K, L$ and the circle $(\omega_2)$ at the point $N$ (with $N$ closer to $L$), then prove that $KC = NL$.