In the parallelogram $CMNP$ extend the bisectors of angles $MCN$ and $PCN$ and intersect with extensions of sides PN and $MN$ at points $A$ and $B$, respectively. Prove that the bisector of the original angle $C$ of the the parallelogram is perpendicular to $AB$. [img]https://cdn.artofproblemsolving.com/attachments/f/3/fde8ef133758e06b1faf8bdd815056173f9233.png[/img]

Let $ABCD$ be a square. Let $M$ and $K$ be points on segments $BC$ and $CD$ respectively, such that $MC = KD$. Let $ P$ be the intersection of the segments $MD$ and $BK$. Prove that $AP$ is perpendicular to $MK$.

Let $ABCD$ be a convex quadrilateral inscibed in circle $(O)$ such that $DB = DA + DC$. The point $P$ lies on the ray $AC$ such that $AP = BC$. The point $E$ is on $(O)$ such that $BE \perp AD$. Prove that $DP$ is parallel to the angle bisector of $\angle BEC$.

Let $ABC$ be a triangle inscribed in a circle of center $O$ and $P$ be a point on the arc $AB$, that does not contain $C$. The perpendicular drawn fom $P$ on line $BO$ intersects $AB$ at $S$ and $BC$ at $T$. The perpendicular drawn from $P$ on line $AO$ intersects $AB$ at $Q$ and $AC$ at $R$. Prove that: a) $PQS$ is an isosceles triangle b) $PQ^2=QR= ST$

A circle $c$ with center $A$ passes through the vertices $B$ and $E$ of a regular pentagon $ABCDE$. The line $BC$ intersects the circle $c$ for second time at point $F$. Prove that the lines $DE$ and $EF$ are perpendicular.

Let $ABCD$ be a cyclic quadrilateral inscribed in circle $O$, where $AC\perp BD$. $M,N$ are the midpoint of arc $ADC,ABC$. $DO$ and $AN$ intersect each other at $G$, the line passes through $G$ and parellel to $NC$ intersect $CD$ at $K$. Prove that $AK\perp BM$.

Let $BD$ and $CE$ be the altitudes of triangle $ABC$ that intersect at point $H$. Let $F$ be a point on side $AC$ such that $FH\perp CE$. The segment $FE$ intersects the circumcircle of triangle $CDE$ at the point $K$. Prove that $HK\perp EF$ . (Matthew Kurskyi)

Points $P$ and $Q$ lies inside triangle $ABC$ such that $\angle ACP =\angle BCQ$ and $\angle CAP = \angle BAQ$. Denote by $D,E$, and $F$ the feet of perpendiculars from $P$ to lines $BC,CA$, and $AB$, respectively. Prove that if $\angle DEF = 90^o$, then $Q$ is the orthocenter of triangle $BDF$.

The chords $AC$ and $BD$ of the circle intersect at point $P$. The perpendiculars to $AC$ and $BD$ at points $C$ and $D$, respectively, intersect at point $Q$. Prove that the lines $AB$ and $PQ$ are perpendicular.

Suppose triangle $ABC$ is right-angled at $B$. The altitude from $B$ intersects the side $AC$ at point $D$. If points $E$ and $F$ are the midpoints of $BD$ and $CD$, prove that $AE \perp BF$.

Circle $S$ with center $O$ and circle $S'$ intersect at points $A$ and $B$. Point $C$ is taken on the arc of a circle $S$ lying inside $S'$. Denote the intersection points of $AC$ and $BC$ with $S'$, other than $A$ and $B$, as $E$ and $D$, respectively. Prove that lines $DE$ and $OC$ are perpendicular.

Let $ABC$ be an acute angled triangle with circumcircle $O$. Line $AO$ intersects the circumcircle of triangle $ABC$ again at point $D$. Let $P$ be a point on the side $BC$. Line passing through $P$ perpendicular to $AP$ intersects lines $DB$ and $DC$ at $E$ and $F$ respectively . Line passing through $D$ perpendicular to $BC$ intersects $EF$ at point $Q$. Prove that $EQ = FQ$ if and only if $BP = CP$.

Let $ABCD$ be a square. On the sides $BC$ and $CD$ the points $M$ and $K$ respectively, so that $MC = KD$. Let $P$ the intersection point of of segments $MD$ and $BK$. Prove that $AP \perp MK$.

Two circle $\Gamma_1$ and $\Gamma_2$ intersect at points $A$ and $B$, point $P$ is not on the line $AB$. Line $AP$ intersects again $\Gamma_1$ and $\Gamma_2$ at points $K$ and $L$ respectively, line $BP$ intersects again $\Gamma_1$ and $\Gamma_2$ at points $M$ and $N$ respectively and all the points mentioned so far are different. The centers of the circles circumscribed around the triangles $KMP$ and $LNP$ are $O_1$ and $O_2$ respectively. Prove that $O_1O_2$ is perpendicular to $AB$.

As shown in figure , it is known that $M$ is the midpoint of side $BC$ of $\vartriangle ABC$. $\odot O$ passes through points $A, C$ and is tangent to $AM$. The extension of the segment $BA$ intersects $\odot O$ at point $D$. The lines $CD$ and $MA$ intersect at the point $P$. Prove that $PO \perp BC$. [img]https://cdn.artofproblemsolving.com/attachments/8/a/da3570ec7eb0833c7a396e22ffac2bd8902186.png[/img]

a) Points $A_1, A_2, A_3, A_4, A_5, A_6$ divide a circle of radius $1$ into six equal arcs. Ray $\ell_1$ from $A_1$ connects $A_1$ with $A_2$, ray $\ell_2$ from $A_2$ connects $A_2$ with $A_3$, and so on, ray $\ell_6$ from $A_6$ connects $A_6$ with $A_1$. From a point $B_1$ on $\ell_1$ the perpendicular is drawn on $\ell_6$, from the foot of this perpendicular another perpendicular is drawn on $\ell_5$, and so on. Let the foot of the $6$-th perpendicular coincide with $B_1$. Find the length of segment $A_1B_1$. b) Find points $B_1, B_2,... , B_n$ on the extensions of sides $A_1A_2, A_2A_3,... , A_nA_1$ of a regular $n$-gon $A_1A_2...A_n$ such that $B_1B_2 \perp A_1A_2$, $B_2B_3 \perp A_2A_3$,$ . . . $, $B_nB_1 \perp A_nA_1$.

Let $ABC$ be a triangle such that $M$ and $N$ are the midpoints of $AC$ and $BC$, respectively. Let $I$ be the incenter of $ABC$ and $E$ be the intersection of $MN$ with $Bl$. Let $P$ be a point such that $EP$ is perpendicular to $MN$ and $NP$ parallel to $IA$. Prove that $IP$ is perpendicular to $BC$.

Let $\omega$ be a circle with center $O$ and let $A$ be a point outside $\omega$. The tangents from $A$ touch $\omega$ at points $B$, and $C$. Let $D$ be the point at which the line $AO$ intersects the circle such that $O$ is between $A$ and $D$. Denote by $X$ the orthogonal projection of $B$ onto $CD$, by $Y$ the midpoint of the segment $BX$ and by $Z$ the second point of intersection of the line $DY$ with $\omega$. Prove that $ZA$ and $ZC$ are perpendicular to each other.

Let $C$ be a point lies outside the circle $(O)$ and $CS, CT$ are tangent lines of $(O)$. Take two points $A, B$ on $(O)$ with $M$ is the midpoint of the minor arc $AB$ such that $A, B, M$ differ from $S, T$. Suppose that $MS, MT$ cut line $AB$ at $E, F$. Take $X \in OS$ and $Y \in OT$ such that $EX, FY$ are perpendicular to $AB$. Prove that $X Y$ and $C M$ are perpendicular.

Consider an acute triangle $ABC$ and it's circumcircle $\omega$. With center $A$, we construct a circle $\gamma$ that intersects arc $AB$ of circle $\omega$ , that doesn't contain $C$, at point $D$ and arc $AC$ , that doesn't contain $B$, at point $E$. Suppose that the intersection point $K$ of lines $BE$ and $CD$ lies on circle $\gamma$. Prove that line $AK$ is perpendicular on line $BC$.

Let $ABC$ be a triangle with $AB \ne AC$. The incirle of triangle $ABC$ is tangent to $BC, CA, AB$ at $D, E, F$, respectively. The perpendicular line from $D$ to $EF$ intersects $AB$ at $X$. The second intersection point of circumcircles of triangles $AEF$ and $ABC$ is $T$. Prove that $TX \perp T F$

$ABCD$ is an isosceles trapezium such that $AD=BC$, $AB=5$ and $CD=10$. A point $E$ on the plane is such that $AE\perp{EC}$ and $BC=EC$. The length of $AE$ can be expressed as $a\sqrt{b}$, where $a$ and $b$ are integers and $b$ is not divisible by any square number other than $1$. Find the value of $(a+b)$.

A triangle $ABC$ with $AB>AC$ is given, $AD$ is the A-angle bisector with point $D$ on $BC$ and point $I$ is the incenter of triangle $ABC$. Point M is the midpoint of segment $AD$ and point $F$ is the second intersection of $MB$ with the circumcirle of triangle $BIC$. Prove that $AF\bot FC$.

As shown in the figure, $BQ$ is a diameter of the circumcircle of $ABC$, and $D$ is the midpoint of arc $BC$ (excluding point $A$) . The bisector of the exterior angle of $\angle BAC$ intersects and the extension of $BC$ at point $E$. The ray $EQ$ intersects $\odot (ABC)$ at point $P$. Point $S$ lies on $PQ$ so that $SA = SP$. Point $T$ lies on $BC$ such that $TB = TD$. Prove that $TS \perp SE$. [img]https://cdn.artofproblemsolving.com/attachments/c/4/01460565e70b32b29cddb65d92e041bea40b25.png[/img]

Given a triangle $ABC$, in which the medians $BE$ and $CF$ are perpendicular. Let $M$ is the intersection point of the medians of this triangle, and $L$ is its Lemoine point (the intersection point of lines symmetrical to the medians with respect to the bisectors of the corresponding angles). Prove that $ML \perp BC$. (Mykhailo Sydorenko)