In an acute triangle $ABC$ let $K,L,$ and $M$ be the midpoints of sides $AB,BC,$ and $CA,$ respectively. From each of $K,L,$ and $M$ drop two perpendiculars to the other two sides of the triangle; e.g., drop perpendiculars from $K$ to sides $BC$ and $CA,$ etc. The resulting $6$ perpendiculars intersect at points $Q,S,$ and $T$ as in the figure to form a hexagon $KQLSMT$ inside triangle $ABC.$ Prove that the area of this hexagon $KQLSMT$ is half of the area of the original triangle $ABC.$ [asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra; diagram by adihaya*/ import graph; size(10cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = 11.888712276357234, xmax = 17.841346447833423, ymin = 10.61620970860601, ymax = 15.470685507068502; /* image dimensions */ pen zzttqq = rgb(0.6,0.2,0.); pen qqwuqq = rgb(0.,0.39215686274509803,0.); pair A = (12.488234161849352,12.833838721895551), B = (16.50823416184936,15.093838721895553), C = (16.28823416184936,11.353838721895551), K = (14.498234161849355,13.963838721895552), L = (16.39823416184936,13.223838721895552), M = (14.388234161849356,12.093838721895551), D = (13.615830174638527,13.467760858438725), F = (15.75135711740064,11.562938202365055), G = (15.625830174638523,14.597760858438724), H = (16.435061748056253,13.849907687412797), T = (14.02296781802369,12.74356027153236), Q = (16.032967818023693,13.873560271532357), O = (16.325061748056253,11.979907687412794); draw(A--B--C--cycle, zzttqq); draw((13.426050287639166,13.361068683160477)--(13.532742462917415,13.171288796161116)--(13.722522349916774,13.277980971439364)--D--cycle, qqwuqq); draw((14.054227993863618,12.223925334689998)--(14.133240861538676,12.426796211152979)--(13.930369985075695,12.505809078828037)--(13.851357117400637,12.302938202365056)--cycle, qqwuqq); draw((16.337846386707046,12.19724654447628)--(16.12050752964356,12.210031183127075)--(16.107722890992765,11.992692326063588)--O--cycle, qqwuqq); draw((15.830369985075697,11.765809078828037)--(15.627499108612716,11.844821946503092)--(15.54848624093766,11.641951070040111)--F--cycle, qqwuqq); draw((15.436050287639164,14.491068683160476)--(15.542742462917412,14.301288796161115)--(15.73252234991677,14.407980971439365)--G--cycle, qqwuqq); draw((16.217722890992764,13.86269232606359)--(16.20493825234197,13.645353469000101)--(16.42227710940546,13.63256883034931)--H--cycle, qqwuqq); Label laxis; laxis.p = fontsize(10); xaxis(xmin, xmax, Ticks(laxis, Step = 1., Size = 2, NoZero),EndArrow(6), above = true); yaxis(ymin, ymax, Ticks(laxis, Step = 1., Size = 2, NoZero),EndArrow(6), above = true); /* draws axes; NoZero hides '0' label */ /* draw figures */ draw(A--B, zzttqq); draw(B--C, zzttqq); draw(C--A, zzttqq); draw(M--D); draw(K--(13.851357117400637,12.302938202365056)); draw(F--L); draw(L--G); draw(K--H); draw(M--O); /* dots and labels */ dot(A,dotstyle); label("$A$", (12.52502834296331,12.93568440300881), NE * labelscalefactor); dot(B,dotstyle); label("$B$", (16.548187989892043,15.193580123223922), NE * labelscalefactor); dot(C,dotstyle); label("$C$", (16.332661580235147,11.457789022504372), NE * labelscalefactor); dot(K,linewidth(3.pt) + dotstyle); label("$K$", (14.536608166427676,14.02357961365791), NE * labelscalefactor); dot(L,linewidth(3.pt) + dotstyle); label("$L$", (16.43529320388129,13.28463192340569), NE * labelscalefactor); dot(M,linewidth(3.pt) + dotstyle); label("$M$", (14.433976542781535,12.155684063298134), NE * labelscalefactor); dot(D,linewidth(3.pt) + dotstyle); dot((13.851357117400637,12.302938202365056),linewidth(3.pt) + dotstyle); dot(F,linewidth(3.pt) + dotstyle); dot(G,linewidth(3.pt) + dotstyle); dot(H,linewidth(3.pt) + dotstyle); dot((15.922967818023695,12.003560271532354),linewidth(3.pt) + dotstyle); label("$S$", (15.96318773510904,12.063315602016607), NE * labelscalefactor); dot(T,linewidth(3.pt) + dotstyle); label("$T$", (14.064502697655428,12.802263292268826), NE * labelscalefactor); dot(Q,linewidth(3.pt) + dotstyle); label("$Q$", (16.076082521119794,13.931211152376383), NE * labelscalefactor); dot(O,linewidth(3.pt) + dotstyle); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy]

Let $M$ and $I$ be the centroid and the incenter of a scalene triangle $ABC$, and let $r$ be its inradius. Prove that $MI = r/3$ if and only if $MI$ is perpendicular to one of the sides of the triangle. (A.Karlyuchenko)

Given an acute-angled triangle $ABC$. Let $CB$ be the altitude and $E$ a random point on the line $CD$. Finally, let $P, Q, R$ and $S$ are the projections of $D$ on the straight lines $AC, AE, BE$ and $BC$. Prove that the points $P, Q, R$ and $S$ lie either on a circle or on one straight line.

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 $CK \perp AB$.

Diagonals of an isosceles trapezoid $ABCD$ with bases $BC$ and $AD$ are perpendicular. Let $DE$ be the perpendicular from $D$ to $AB$, and let $CF$ be the perpendicular from $C$ to $DE$. Prove that angle $DBF$ is equal to half of angle $FCD$.

The quadrangle $ABCD$ is inscribed in a circle whose center $O$ lies inside it. Prove that if $\angle BAO = \angle DAC$, then the diagonals of the quadrilateral are perpendicular.

Let $ABC$ be an acute-angled triangle and let $P$ be the intersection of the tangents at $B$ and $C$ of the circumscribed circle of $\vartriangle ABC$. The line through $A$ perpendicular on $AB$ and cuts the line perpendicular on $AC$ through $C$ at $X$. The line through $A$ perpendicular on $AC$ cuts the line perpendicular on $AB$ through $B$ at $Y$. Show that $AP \perp XY$.

Let be the regular hexagonal prism $ABCDEFA'B C'D'E'F'$ with the base edge of $12$ and the height of $12 \sqrt{3}$. We denote by $N$ the middle of the edge $CC'$. a) Prove that the lines $BF'$ and $ND$ are perpendicular b) Calculate the distance between the lines $BF'$ and $ND$.

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 $HE = FG$.

Given an equilateral $\Delta ABC$, in which ${{A} _ {1}}, {{B} _ {1}}, {{C} _ {1}}$ are the midpoint of the sides $ BC, \, \, AC, \, \, AB$ respectively. The line $l$ passes through the vertex $A$, we denote by $P, Q$ the projection of the points $B, C$ on the line $l$, respectively (the line $ l $ and the point $Q, \, \, A, \, \, P$ are located as shown in fig.). Denote by $T $ the intersection point of the lines ${{B} _ {1}} P$ and ${{C} _ {1}} Q$. Prove that the line ${{A} _ {1}} T$ is perpendicular to the line $l$. [img]https://cdn.artofproblemsolving.com/attachments/4/b/61f2f4ec9e6b290dfcd47e9351110bebd3bd43.png[/img] (Serdyuk Nazar)

Two circles $\omega_1$ and $\omega_2$ are tangent to line $\ell$ at the points $A$ and $B$ respectively. In addition, $\omega_1$ and $\omega_2 $are externally tangent to each other at point $D$. Choose a point $E$ on the smaller arc $BD$ of circle $\omega_2$. Line $DE$ intersects circle $\omega_1$ again at point $C$. Prove that $BE \perp AC$. (Yurii Biletskyi)

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]

Let the isosceles triangle $ABC$ with $| AB | = | AC |$. The point $M$ is the midpoint of the base $[BC]$, the point $N$ is the orthogonal projection of the point $M$ on the line $AC$, and the point $P$ is located on the segment $(MC)$ such that $| MP | = | P C | \sin^2 C$. Prove that the lines $AP$ and $BN$ are perpendicular.

Outside the square $ABCD$, the rhombus $BCMN$ is constructed with angle $BCM$ obtuse . Let $P$ be the intersection point of the lines $BM$ and $AN$ . Prove that $DM \perp CP$ and the triangle $DPM$ is right isosceles .

In a triangle $ABC$, the bisector of the largest angle $\angle A$ meets $BC$ at point $D$. Let $E$ and $F$ be the feet of perpendiculars from $D$ to $AC$ and $AB$, respectively. Let $R$ denote the ratio between the areas of triangles $DEB$ and $DFC$. (a) Prove that, for every real number $r > 0$, one can construct a triangle ABC for which $R$ is equal to $r$. (b) Prove that if $R$ is irrational, then at least one side length of $\vartriangle ABC$ is irrational. (c) Give an example of a triangle $ABC$ with exactly two sides of irrational length, but with rational $R$.

In triangle $ABC$, the points $D$, $E$, and $F$ are the feet of the perpendiculars dropped from $A$, $B$, and $C$, respectively, onto the opposite sides. The point $X_A$ is such that a circle passing through $E$ and $F$ is tangent to the circumcircle of triangle $ABC$ at $X_A$, and $X_A$ is on a different side of $EF$ as $A$. Similarly, $X_B$ and $X_C$ are defined. Prove that the lines $AX_A$, $BX_B$, and $CX_C$ are concurrent.

Prove that the edges $AB$ and $CD$ of a tetrahedron $ABCD$ are perpendicular if and only if there exists a parallelogram $CDPQ$ such that $PA = PB = PD$ and $QA = QB = QC$.

As shown in the figure, it is known that the quadrilateral $ABCD$ satisfies $\angle ADB = \angle ACB = 90^o$. Suppose $AC$ and $BD$ intersect at point $P$, point $R$ lies on $CD$ and $RP \perp AB$. $M$ and $N$ are the midpoints of $AB$ and $CD$ respectively. Point $K$ is a point on the extension line of $NM$, the circumscribed circles of $\vartriangle DKC$ and $\vartriangle AKB$ intersect at point $S$. Prove that $KS \perp SR$. [img]https://cdn.artofproblemsolving.com/attachments/5/d/fc0a391f8ebcdee792e9b226cbf55a058251a1.png[/img]

Let $ABCD$ ba a square and let point $E$ be the midpoint of side $AD$. Points $G$ and $F$ are located on the segment $(BE)$ such that the lines $AG$ and $CF$ are perpendicular on the line $BE$. Prove that $DF= CG$.

A circle intersects a parabola at four distinct points. Let $M$ and $N$ be the midpoints of the arcs of the circle which are outside the parabola. Prove that the line $MN$ is perpendicular to the axis of the parabola. I. Voronovich

In the convex quadrilateral of the $ABCD$, the diagonals of $AC$ and $BD$ are mutually perpendicular and intersect at point $E$. On the side of $AD$, a point $P$ is chosen such that $PE = EC$. The circumscribed circle of the triangle $BCD$ intersects the segment $AD$ at the point $Q$. The circle passing through point $A$ and tangent to the line $EP$ at point $P$ intersects the segment $AC$ at point $R$. It turns out that points $B, Q, R$ are collinear. Prove that $\angle BCD = 90^o$.

Given is a triangle with side lengths $a,b$ and $c$, incenter $I$ and centroid $S$. Prove: If $a+b=3c$, then $S \neq I$ and line $SI$ is perpendicular to one of the sides of the triangle.

Let $ABCDEFGHI$ be a regular polygon of $9$ sides. The segments $AE$ and $DF$ intersect at $P$. Prove that $PG$ and $AF$ are perpendicular.

Given a triangle $ABC$ and a point $P_0$ on the side $AB$. Construct points $P_i, Q_i, R_i $ as follows. $Q_i$ is the foot of the perpendicular from $P_i$ to $BC, R_i$ is the foot of the perpendicular from $Q_i$ to $AC$ and $P_i$ is the foot of the perpendicular from $R_{i-1}$ to $AB$. Show that the points $P_i$ converge to a point $P$ on $AB$ and show how to construct $P$.

Let $ABCD$ be a parallelogram ($AB < BC$). The bisector of the angle $BAD$ intersects the side $BC$ at the point K; and the bisector of the angle $ADC$ intersects the diagonal $AC$ at the point $F$. Suppose that $KD \perp BC$. Prove that $KF \perp BD$.