Let $ACE$ be a triangle with $\angle ECA=60^{\circ}, \angle AEC=90^{\circ}$. Let $B$ and $D$ be points on the sides $AC$ and $CE$ respectively such that the $\triangle BCD$ is equilateral. Now suppose $BD \cap AE=F$. Find $\angle EAC+\angle EFD$.

In a right-angled and isosceles triangle, the two catheti are both length $1$. Find the length of the shortest line segment dividing the triangle into two figures with the same area, and specify the location of this line segment

Let $XOY$ be a triangle with $\angle XOY = 90^o$. Let $M$ and $N$ be the midpoints of legs $OX$ and $OY$, respectively. Suppose that $XN = 19$ and $YM =22$. What is $XY$?

Consider a right -angle triangle $ABC$ with $A=90^{o}$, $AB=c$ and $AC=b$. Let $P\in AC$ and $Q\in AB$ such that $\angle APQ=\angle ABC$ and $\angle AQP = \angle ACB$. Calculate $PQ+PE+QF$, where $E$ and $F$ are the projections of $B$ and $Q$ onto $BC$, respectively.

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]

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 $\angle ACB = 90^{\circ}$.

The areas of two right-angled triangles have ratio equal to the squares of their hypotenuses. Show that these triangles are similar.

Given a right triangle $ABC$ with the hypotenuse $AB$. Let $K$ be any interior point of triangle $ABC$ and points $L, M$ are symmetric of point $K$ wrt lines $BC, AC$ respectively. Specify all possible values for $S_{ABLM} / S_{ABC}$, where $S_{XY ... Z}$ indicates the area of the polygon $XY...Z$ .

Let $ABC$ ne a right triangle with $\angle ACB=90^o$. Let $E, F$ be respecitvely the midpoints of the $BC, AC$ and $CD$ be it's altitude. Next, let $P$ be the intersection of the internal angle bisector from $A$ and the line $EF$. Prove that $P$ is the center of the circle inscribed in the triangle $CDE$ .

In a right-angled triangle, the sum $a + b$ of the sides enclosing the right angle equals $24$ while the length of the altitude $h_c$ on the hypotenuse $c$ is $7$. Determine the length of the hypotenuse.

On the sides $AB$ and $AC$ of the right $\triangle ABC$ ($\angle A=90^{\circ}$) are chosen points $C_{1}$ and $B_{1}$ respectively. Prove that if $M=CC_{1}\cap BB_{1}$ and $AC_{1}=AB_{1}=AM$, then $[AB_{1}MC_{1}]+[AB_{1}C_{1}]=[BMC]$.

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 $ABC$?

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]

A rectangle, whose one sidelength is twice the other side, is inscribed inside a triangles with sides $3$ cm, $4$ cm and $5$ cm, such that the long sides lies entirely on the long side of the triangle. The other two remaining vertices of the rectangle lie respectively on the other two sides of the triangle. Find the lengths of the sides of this rectangle.

It is known that in a right triangle: a) The height drawn from the top of the right angle is the geometric mean of the projections of the legs on the hypotenuse; b) the leg is the geometric mean of the hypotenuse and the projection of this leg to the hypotenuse. Are the converse statements true? Formulate them and justify the answer. Is it possible to formulate the criterion of a right triangle based on these statements? If possible, then how? If not, why?

In a right-angled triangle, the sum $a + b$ of the sides enclosing the right angle equals $24$ while the length of the altitude $h_c$ on the hypotenuse $c$ is $7$. Determine the length of the hypotenuse.

Let $\triangle ABC$ be a right triangle at the vertex $A$ such that the side $AB$ is shorter than the side $AC$. Let $D$ be the foot of the altitude from $A$ to $BC$ and $M$ the midpoint of $BC$. Let $E$ be a point on the ray $AB$, outside of the segment $AB$. Line $ED$ intersects the segment $AM$ at the point $F$. Point $H$ is on the side $AC$ such that $\angle EFH=90^{\circ}$. Suppose that $ED=FH$. Find the measure of the angle $\angle AED$.

A right-angled triangle made of paper is folded along a straight line so that the vertex at the right angle coincides with one of the other vertices of the triangle and a quadrilateral is obtained . (a) What is the ratio into which the diagonals of this quadrilateral divide each other? (b) This quadrilateral is cut along its longest diagonal. Find the area of the smallest piece of paper thus obtained if the area of the original triangle is $1$ . (A Shapovalov)

The two tangents to the incircle of a right-angled triangle $ABC$ that are perpendicular to the hypotenuse $AB$ intersect it at the points $P$ and $Q$. Find $\angle PCQ$. (M Evdokimov,)

In a right-angled triangle in which all side lengths are integers, one has a cathetus length $1994$. Determine the length of the hypotenuse.

Let $C$ be a right angle in triangle $ABC$. On legs $AC$ and$BC$ the squares $ACKL, BCMN$ are constructed outside of triangle. If $CE$ is an altitude of the triangle prove that $LEM$ is a right angle.

Two circles $\Gamma_a$ and $\Gamma_b$ with their centres lying on the legs $BC$ and $CA$ of a right triangle, both touching the hypotenuse $AB$, and both passing through the vertex $C$ are given. Let the radii of these circles be denoted by $\gamma_a$ and $\gamma_b$. Find the greatest real number $p$ such that the inequality $\frac{1}{\gamma_a}+\frac{1}{\gamma_b}\ge p \left(\frac{1}{a}+\frac{1}{b}\right)$ ($BC = a,CA = b$) holds for all right triangles $ABC$.

Let $ABC$ be a triangle with $\angle CAB > 45^o$ and $\angle CBA > 45^o$. Construct an isosceles right angled triangle $RAB$ with $AB$ as its hypotenuse and $R$ inside $ABC$. Also construct isosceles right angled triangles $ACQ$ and $BCP$ having $AC$ and $BC$ respectively as their hypotenuses and lying entirely outside $ABC$. Show that $CQRP$ is a parallelogram.

An isosceles triangle $ABC$ is inscribed in a circle with $\angle ACB = 90^o$ and $EF$ is a chord of the circle such that neither E nor $F$ coincide with $C$. Lines $CE$ and $CF$ meet $AB$ at $D$ and $G$ respectively. Prove that $|CE|\cdot |DG| = |EF| \cdot |CG|$.

$\vartriangle ABC$ has $AB = 5$, $BC = 12$, and $AC = 13$. A circle is inscribed in $\vartriangle ABC$, and $MN$ tangent to the circle is drawn such that $M$ is on $\overline{AC}$, $N$ is on $\overline{BC}$, and $\overline{MN} \parallel \overline{AB}$. The area of $\vartriangle MNC$ is $m/n$ , where $m$ and $n $are relatively prime positive integers. Find $m + n$.