In the quadrilateral $ABCD$, $AB = BC = CD$ and $\angle BMC = 90^\circ$, where $M$ is the midpoint of $AD$. Determine the acute angle between the lines $AC$ and $BD$.

Let $ABCD$ be a cyclic quadrilateral. Let $H_{1}$ be the orthocentre of triangle $ABC$. Point $A_{1}$ is the image of $A$ after reflection about $BH_{1}$. Point $B_{1}$ is the image of of $B$ after reflection about $AH_{1}$. Let $O_{1}$ be the circumcentre of $(A_{1}B_{1}H_{1})$. Let $H_{2}$ be the orthocentre of triangle $ABD$. Point $A_{2}$ is the image of $A$ after reflection about $BH_{2}$. Point $B_{2}$ is the image of of $B$ after reflection about $AH_{2}$. Let $O_{2}$ be the circumcentre of $(A_{2}B_{2}H_{2})$. Lets denote by $\ell_{AB}$ be the line through $O_{1}$ and $O_{2}$. $\ell_{AD}$ ,$\ell_{BC}$ ,$\ell_{CD}$ are defined analogously. Let $M=\ell_{AB} \cap \ell_{BC}$, $N=\ell_{BC} \cap \ell_{CD}$, $P=\ell_{CD} \cap \ell_{AD}$,$Q=\ell_{AD} \cap \ell_{AB}$. Prove that $MNPQ$ is cyclic.

An acute-angled $ABC \ (AB<AC)$ is inscribed into a circle $\omega$. Let $M$ be the centroid of $ABC$, and let $AH$ be an altitude of this triangle. A ray $MH$ meets $\omega$ at $A'$. Prove that the circumcircle of the triangle $A'HB$ is tangent to $AB$. [i](A.I. Golovanov , A.Yakubov)[/i]

Let $ABC$ be a scalene triangle. The inscribed circle of $ABC$ touches the sides $BC$, $CA$, and $AB$ at the points $A_1$, $B_1$, $C_1$ respectively. Let $I$ be the incenter, $O$ be the circumcenter, and lines $OI$ and $BC$ meet at point $D$. The perpendicular line from $A_1$ to $B_1 C_1$ intersects $AD$ at point $E$. Prove that $B_1 C_1$ passes through the midpoint of $EA_1$.

$ \Delta ABC$ is a triangle such that $ AB \neq AC$. The incircle of $ \Delta ABC$ touches $ BC, CA, AB$ at $ D, E, F$ respectively. $ H$ is a point on the segment $ EF$ such that $ DH \bot EF$. Suppose $ AH \bot BC$, prove that $ H$ is the orthocentre of $ \Delta ABC$. Remark: the original question has missed the condition $ AB \neq AC$

Let $ABC$ be a triangle. Points $D$ and $E$ are taken on the line $BC$ such that $AD$ and $AE$ are parallel to the respective tangents to the circumcircle at $C$ and $B$. Prove that \[\frac{BE}{CD}=\left(\frac{AB}{AC}\right)^2 \]

Let $A$ be a point outside the circle $\omega$ . The tangents from $A$ touch the circle at $B$ and $C$. Let $P$ be an arbitrary point on extension of $AC$ towards $C$, $Q$ the projection of $C$ onto $PB$ and $E$ the second intersection point of the circumcircle of $ABP$ with the circle $\omega$ . Prove that $\angle PEQ = 2\angle APB$

Let $ ABCD$ be a trapezoid such that $ AB \parallel CD$ and assume that there are points $ E$ on the line outside the segment $ BC$ and $ F$ on the segment $ AD$ such that $ \angle DAE \equal{} \angle CBF$. Let $ I,J,K$ respectively be the intersection of line $ EF$ and line $ CD$, the intersection of line $ EF$ and line $ AB$, and the midpoint of segment $ EF$. Prove that $ K$ is on the circumcircle of triangle $ CDJ$ if and only if $ I$ is on the circumcircle of triangle $ ABK$. [i]Utari Wijayanti, Bandung[/i]

In the pyramid $SA_1A_2 \cdots A_n$, all sides are equal. Let point $X_i$ be the midpoint of arc $A_iA_{i+1}$ in the circumcircle of $\triangle SA_iA_{i+1}$ for $1 \le i \le n$ with indices taken mod $n$. Prove that the circumcircles of $X_1A_2X_2, X_2A_3X_3, \cdots, X_nA_1X_1$ have a common point.

Points $A_1,~ B_1,~ C_1$ lie on the sides $BC,~ AC$ and $AB$ of a triangle $ABC$, respectively, such that $AB_1 -AC_1 = CA_1 -CB_1 = BC_1 -BA_1$. Let $I_A,~ I_B,~ I_C$ be the incenters of triangles $AB_1C_1,~ A_1BC_1$ and $A_1B_1C$ respectively. Prove that the circumcenter of triangle $I_AI_BI_C$, is the incenter of triangle $ABC$.

Let $ABC$ be an acute triangle with orthocenter $H$, circumcenter $O$, and circumcircle $\Omega$. Points $E$ and $F$ are the feet of the altitudes from $B$ to $AC$, and from $C$ to $AB$, respectively. Let line $AH$ intersect $\Omega$ again at $D$. The circumcircle of $DEF$ intersects $\Omega$ again at $X$, and $AX$ intersects $BC$ at $I$. The circumcircle of $IEF$ intersects $BC$ again at $G$. If $M$ is the midpoint of $BC$, prove that lines $MX$ and $OG$ intersect on $\Omega$.

Consider $\triangle ABC$ and a point $M$ in its interior so that $\angle MAB = 10^\circ$, $\angle MBA = 20^\circ$, $\angle MCA = 30^\circ$ and $\angle MAC = 40^\circ$. What is $\angle MBC$?

The points $P$ and $Q$ lie on the diagonals $AC$ and $BD$, respectively, of a quadrilateral $ABCD$ such that $\frac{AP}{AC} + \frac{BQ}{BD} =1$. The line $PQ$ meets the sides $AD$ and $BC$ at points $M$ and $N$. Prove that the circumcircles of the triangles $AMP$, $BNQ$, $DMQ$, and $CNP$ are concurrent.

Given a cyclic quadriteral $ABCD$. The circumcenter lies in the quadriteral $ABCD$. Diagonals $AC$ and $BD$ intersects at $S$. Points $P$ and $Q$ are the midpoints of $AD$ and $BC$. Let $p$ be a line perpendicular to $AC$ through $P$, $q$ perpendicular line to $BD$ through $Q$ and $s$ perpendicular to $CD$ through $S$. Prove that $p,q,s$ intersects at one point.

Point $I_a$ is the $A$-excircle center of $\triangle ABC$ which is tangent to $BC$ at $X$. Let $A'$ be diametrically opposite point of $A$ with respect to the circumcircle of $\triangle ABC$. On the segments $I_aX, BA'$ and $CA'$ are chosen respectively points $Y,Z$ and $T$ such that $I_aY=BZ=CT=r$ where $r$ is the inradius of $\triangle ABC$. Prove that the points $X,Y,Z$ and $T$ are concyclic.

In an acute triangle $ABC$ point $D$ is the point of intersection of altitude $h_a$ and side $BC$, and points $M, N$ are orthogonal projections of point $D$ on sides $AB$ and $AC$. Lines $MN$ and $AD$ cross the circumcircle of triangle $ABC$ at points $P, Q$ and $A, R$. Prove that point $D$ is the center of the incircle of $PQR$.

Let $P$, $Q$, $R$ be points on the sides $AB$, $BC$, $CA$ of a triangle $ABC$ such that $AP=CQ$ and the quadrilateral $RPBQ$ is cyclic. The tangents to the circumcircle of triangle $ABC$ at the points $C$ and $A$ intersect the lines $RQ$ and $RP$ at the points $X$ and $Y$, respectively. Prove that $RX=RY$.

A; B; C; D are the vertices of a square, and P is a point on the arc CD of its circumcircle. Prove that $ |PA|^2 - |PB|^2 = |PB|.|PD| -|PA|.|PC| $ Can anyone here find the solution? I'm not great with geometry, so i tried turning it into co-ordinate geometry equations, but sadly to no avail. Thanks in advance.

A non-equilateral triangle $ABC$ is inscribed in a circle $\Gamma$ with centre $O$, radius $R$ and its incircle has centre $I$ and touches $BC,CA,AB$ at $D,E, F$, respectively. A circle with centre $I$ and radius $r$ intersects the rays $[ID), [IE), [IF)$ at $A',B',C'$. Show that the orthocentre $K$ of $\vartriangle A'B'C'$ is on the line $OI$ and that $\frac{IK}{IO}=\frac{r}{R}$ Michel Bataille .

Let $ A$ be one of the two points of intersection of two circles with centers $ X, Y$ respectively.The tangents at $ A$ to the two circles meet the circles again at $ B, C$. Let a point $ P$ be located so that $ PXAY$ is a parallelogram. Show that $ P$ is also the circumcenter of triangle $ ABC$.

In acute-angled triangle $ABC$, the height $AH$ and median $BM$ were drawn. Point $D$ lies on the circumcircle of triangle $BHM$ such that $AD \parallel BM$ and $B, D$ are on opposite sides of line $AC$. Prove that $BC=BD$.

Let $ABCD$ be a curcumscribed quadrilateral with incenter $I$, and let $O_{1}, O_{2}$ be the circumcenters of triangles $AID$ and $CID$. Prove that the circumcenter of triangle $O_{1}IO_{2}$ lies on the bisector of angle $ABC$

We are given triangles $ABC$ and $DEF$ such that $D\in BC, E\in CA, F\in AB$, $AD\perp EF, BE\perp FD, CF\perp DE$. Let the circumcenter of $DEF$ be $O$, and let the circumcircle of $DEF$ intersect $BC,CA,AB$ again at $R,S,T$ respectively. Prove that the perpendiculars to $BC,CA,AB$ through $D,E,F$ respectively intersect at a point $X$, and the lines $AR,BS,CT$ intersect at a point $Y$, such that $O,X,Y$ are collinear. [i]Proposed by Sammy Luo[/i]

Let $ABCD$ be a convex quadrilateral with $\angle{DAC}= \angle{BDC}= 36^\circ$ , $\angle{CBD}= 18^\circ$ and $\angle{BAC}= 72^\circ$. The diagonals and intersect at point $P$ . Determine the measure of $\angle{APD}$.

A triangle $ABC$ and a point $P$ inside it are given. $A', B', C'$ are the projections of $P$ onto the straight lines ot the sides $BC,CA,AB$. Prove that the center of the circle circumscribed around the triangle $A'B'C'$ lies inside the triangle $ABC$.