In a right triangle $ABC$ with $\angle BAC =90^o $and $\angle ABC= 54^o$, point $M$ is the midpoint of the hypotenuse $[BC]$ , point $D$ is the foot of the angle bisector drawn from the vertex $C$ and $AM \cap CD = \{E\}$. Prove that $AB= CE$.

The triangle $ABC$ is right in $A$ and $R$ is the midpoint of the hypotenuse $BC$ . On the major leg $AB$ the point $P$ is marked such that $CP = BP$ and on the segment $BP$ the point $Q$ is marked such that the triangle $PQR$ is equilateral. If the area of triangle $ABC$ is $27$, calculate the area of triangle $PQR$ .

Triangle $ABC$ with $AB = 10$ cm ¸and $\angle C= 15^o$, is right at $B$. Point $D \in (AC)$ is the foot of the altitude taken from $B$. Find the distance from point $D$ to the line $AB$.

On the hypotenuse $AB$ of a right triangle $ABC$, we denote a point $K$ such that $BK = BC$. Let $P$ be a point on the perpendicular from the point $K$ to line $CK$, equidistant from the points $K$ and $B$. Let $L$ be the midpoint of $CK$. Prove that line $AP$ is tangent to the circumcircle of $\Delta BLP$.

Let $ABC$ be a triangle with $\angle BCA=90^{\circ}$, and let $D$ be the foot of the altitude from $C$. Let $X$ be a point in the interior of the segment $CD$. Let $K$ be the point on the segment $AX$ such that $BK=BC$. Similarly, let $L$ be the point on the segment $BX$ such that $AL=AC$. Let $M$ be the point of intersection of $AL$ and $BK$. Show that $MK=ML$. [i]Proposed by Josef Tkadlec, Czech Republic[/i]

The square $ABCD$ is divided into $8$ equal right triangles and the square $KLMN$, as shown in the figure. Find the area of the square $ABCD$ if $KL = 5, PS = 8$. [img]https://1.bp.blogspot.com/-B2QIHvPcIx0/X4BhUTMDhSI/AAAAAAAAMj4/4h0_q1P6drskc5zSvtfTZUskarJjRp5LgCLcBGAsYHQ/s0/Yasinsky%2B2020%2Bp1.png[/img]

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]

(i) $ABC$ is a triangle with a right angle at $A$, and $P$ is a point on the hypotenuse $BC$. The line $AP$ produced beyond $P$ meets the line through $B$ which is perpendicular to $BC$ at $U$. Prove that $BU = BA$ if, and only if, $CP = CA$. (ii) $A$ is a point on the semicircle $CB$, and points $X$ and $Y$ are on the line segment $BC$. The line $AX$, produced beyond $X$, meets the line through $B$ which is perpendicular to $BC$ at $U$. Also the line $AY$, produced beyond $Y$, meets the line through $C$ which is perpendicular to $BC$ at $V$. Given that $BY = BA$ and $CX = CA$, determine the angle $\angle VAU$.

In the triangle $ABC$, $\angle A=90^\circ$, the bisector of $\angle B$ meets the altitude $AD$ at the point $E$, and the bisector of $\angle CAD$ meets the side $CD$ at $F$. The line through $F$ perpendicular to $BC$ intersects $AC$ at $G$. Prove that $B,E,G$ are collinear.

Consider the isosceles right triangle$ ABC, \angle A = 90^o$, and point $D \in (AB)$ such that $AD = \frac13 AB$. In the half-plane determined by the line $AB$ and point $C$ , consider a point $E$ such that $\angle BDE = 60^o$ and $\angle DBE = 75^o$. Lines $BC$ and $DE$ intersect at point $G$, and the line passing through point $G$ parallel to the line $AC$ intersects the line $BE$ at point $H$. Prove that the triangle $CEH$ is equilateral.

Let $\vartriangle ABC$ be a right triangle with $\angle B = 90^o$. Let $P$ be a point on side $BC$, and let $\omega$ be the circle with diameter $CP$. Suppose that $\omega$ intersects $AC $and $AP$ again at $Q$ and $R$, respectively. Show that $CP^2 = AC \cdot CQ - AP \cdot P R$.

Each leg of a right triangle is increased by one. Could its hypotenuse increase by more than $\sqrt2$?

Let \( I \) and \( O \) be the incenter and circumcenter of the right triangle \( ABC \) (\( \angle C = 90^\circ \)), and let \( K \) be the tangency point of the incircle with \( AC \). Let \( P \) and \( Q \) be the points where the circumcircle of triangle \( AOK \) intersects \( OC \) and the circumcircle of triangle \( ABC \), respectively. Prove that points \( C, I, P, \) and \( Q \) are concyclic. [i]Proposed by Mykhailo Sydorenko[/i]

Let $ABC$ be a right triangle in $A$ , whose leg measures $1$ cm. The bisector of the angle $BAC$ cuts the hypotenuse in $R$, the perpendicular to $AR$ on $R$ , cuts the side $AB$ at its midpoint. Find the measurement of the side $AB$ .

Inside an isosceles right triangle $ABC$ with hypotenuse $AB$ a point $M$ is taken such that the angle $\angle MAB$ is $15 ^o$ larger than the angle $\angle MAC$ , and the angle $\angle MCB$ is $15^o$ larger than the angle $\angle MBC$. Find the angle $\angle BMC$ .

Let $ABC$ be an isosceles triangle with $BC=CA$, and let $D$ be a point inside side $AB$ such that $AD< DB$. Let $P$ and $Q$ be two points inside sides $BC$ and $CA$, respectively, such that $\angle DPB = \angle DQA = 90^{\circ}$. Let the perpendicular bisector of $PQ$ meet line segment $CQ$ at $E$, and let the circumcircles of triangles $ABC$ and $CPQ$ meet again at point $F$, different from $C$. Suppose that $P$, $E$, $F$ are collinear. Prove that $\angle ACB = 90^{\circ}$.

In the Triangle ABC AB>AC, the extension of the altitude AD with D lying inside BC intersects the circum-circle of the Triangle ABC at P. The circle through P and tangent to BC at D intersects the circum-circle of Triangle ABC at Q distinct from P with PQ=DQ. Prove that AD=BD-DC

Points $E$ and $F$ are chosen on the sides $AB$ and $BC$ respectively of a square $ABCD$ such that $BE=BF$. Let $BN$ be an altitude of the triangle $BCE$. Prove that the triangle $DNF$ is right-angled.

Given a right angled triangle $ABC$ in which $\angle ACB = 90^o$. Let $D$ be an inner point of $AB$, and let $E$ be an inner point of $AC$. It is known that $\angle ADE = 90^o$, and that the length of the segment $AD$ is $8$, the length of the segment $DE$ is $15$, and the length of segment $CE$ is $3$. What is the area of triangle $ABC$?

Triangle $ABC_0$ has a right angle at $C_0$. Its side lengths are pairwise relatively prime positive integers, and its perimeter is $p$. Let $C_1$ be the foot of the altitude to $\overline{AB}$, and for $n\geq 2$, let $C_n$ be the foot of the altitude to $\overline{C_{n-2}B}$ in $\triangle C_{n-2}C_{n-1}B$. The sum $\sum\limits_{n=1}^{\infty}C_{n-1}C_n = 6p$. Find $p$.

Consider $\vartriangle ABC$, right at $B$, let $I$ be its incenter and $F,D,E$ the points where the circle inscribed on sides AB, $BC$ and $AC$, respectively. If $M$ is the intersection point of $CI$ and $EF$, and $N$ is the intersection point of $DM$ and $AB$. Prove that $AN = ID$.

The right triangles $ABC$ and $AB_1C_1$ are similar and have opposite orientation. The right angles are at $C$ and $C_1$, and we also have $ \angle CAB = \angle C_1AB_1$. Let $M$ be the point of intersection of the lines $BC_1$ and $B_1C$. Prove that if the lines $AM$ and $CC_1$ exist, they are perpendicular.

In the exterior of the triangle $ABC$ with $m(\angle B) > 45^o$, $m(\angle C) >45°^o$ one constructs the right isosceles triangles $ACM$ and $ABN$ such that $m(\angle CAM) = m(\angle BAN) = 90^o$ and, in the interior of $ABC$, the right isosceles triangle $BCP$, with $m(\angle P) = 90^o$. Show that the triangle $MNP$ is a right isosceles triangle.

Let $ABC$ be a triangle with $\angle A = 90^o$ and $\angle B < \angle C$. The tangent at $A$ to the circumcircle $k$ of $\vartriangle ABC$ intersects line $BC$ at $D$. Let $E$ be the reflection of $A$ in $BC$. Also, let $X$ be the feet of the perpendicular from $A$ to $BE$ and let $Y$ be the midpoint of $AX$. Line $BY$ meets $k$ again at $Z$. Prove that line $BD$ is tangent to the circumcircle of $\vartriangle ADZ$.

Consider a parallelogram $ABCD$ such that the midpoint $M$ of the side $CD$ lies on the angle bisector of $\angle BAD$. Show that $\angle AMB$ is a right angle.