A semicircle $S$ is drawn on $AB$ as diameter. For an arbitrary point $C$ in $S$ ($C\ne A$,$ C \ne B$), squares are attached to sides $AC$ and $BC$ of triangle $ABC$ outside the triangle. Find the locus of the midpoint of the segment joining the centres of the squares as $C$ moves along $S$. (J Tabov, Sofia)

Let be given points $A,B,C$ on a line, with $C$ between $A$ and $B$. Three semicircles with diameters $AC,BC,AB$ are drawn on the same side of line $ABC$. The perpendicular to $AB$ at $C$ meets the circle with diameter $AB$ at $H$. Given that $CH =\sqrt2$, compute the area of the region bounded by the three semicircles.

Let the points $A,B,C$ lie on a line in this order. $AB$ is the diameter of semicircle $\omega_1$, $AC$ is the diameter of semicircle $\omega_2$. Assume both $\omega_1$ and $\omega_2$ lie on the same side of $AC$. $D$ is a point on $\omega_2$ such that $BD\perp AC$. A circle centered at $B$ with radius $BD$ intersects $\omega_1$ at $E$. $F$ is on $AC$ such that $EF\perp AC$. Prove that $BC=BF$.

Let $\vartriangle ABC$ be a right triangle with $\angle B = 90^o$ and $|AB| > |BC|$, and let $\Gamma$ be the semicircle with diameter $AB$ that lies on the same side as $C$. Let $P$ be a point on $\Gamma$ such that $|BP| = |BC|$ and let $Q$ be on $AB$ such that $|AP| = |AQ|$. Prove that the midpoint of $CQ$ lies on $\Gamma$.

The figure shows a rectangle, its circumscribed circle and four semicircles, which have the rectangle’s sides as diameters. Prove that the combined area of the four dark gray crescentshaped regions is equal to the area of the light gray rectangle. [img]https://1.bp.blogspot.com/-gojv6KfBC9I/XzT9ZMKrIeI/AAAAAAAAMVU/NB-vUldjULI7jvqiFWmBC_Sd8QFtwrc7wCLcBGAsYHQ/s0/2013%2BMohr%2Bp3.png[/img]

In a semicircle of radius $4$ three circles are inscribed, as indicated in the figure. Larger circles have radii $ R_1 $ and $ R_2 $, and the larger circle has radius $ r $. a) Prove that $ \dfrac {1} {\sqrt{r}} = \dfrac {1} {\sqrt{R_1}} + \dfrac {1} {\sqrt{R_2}} $ b) Prove that $ R_1 + R_2 \le 8 (\sqrt{2} -1) $ c) Prove that $ r \le \sqrt{2} -1 $ [img]https://cdn.artofproblemsolving.com/attachments/0/9/aaaa65d1f4da4883973751e1363df804b9944c.jpg[/img]

Above the segments $AB$ and $BC$ we drew a semicircle at each. $F_1$ bisects $AB$ and $F_2$ bisects $BC$. Above the segments $AF_2$ and $F_1C$ we also drew a semicircle at each. Segments $P Q$ and $RS$ touch the corresponding semicircles as shown in the figure. Prove that $P Q \parallel RS$ and $|P Q| = 2 \cdot |RS|$. [img]https://cdn.artofproblemsolving.com/attachments/8/2/570e923b91e9e630e3880a014cc6df4dc33aa2.png[/img]

Given a semicirle with center $O$ an arbitrary inner point of the diameter divides it into two segments. Let there be semicircles above the two segments as visible in the below figure. The line $\ell$ passing through the point $A$ intersects the semicircles in $4$ points: $B, C, D$ and $E$. Show that the segments $BC$ and $DE$ have the same length. [img]https://cdn.artofproblemsolving.com/attachments/1/4/86a369d54fef7e25a51fea6481c0b5e7dd45ff.png[/img]

Given a semicircle of diameter $AB$ and center $O$. Let $CD$ be the chord of the semicircle tangent to two circles of diameters $AO$ and $OB$. If $CD=120$ cm,, caclulate area of the semicircle.

A semicircular paper is folded along a chord such that the folded circular arc is tangent to the diameter of the semicircle. The radius of the semicircle is $4$ units and the point of tangency divides the diameter in the ratio $7 :1$. If the length of the crease (the dotted line segment in the figure) is $\ell$ then determine $ \ell^2$. [img]https://cdn.artofproblemsolving.com/attachments/5/6/63fed83742c8baa92d9e63962a77a57d43556f.png[/img]

Given semicircle $(c)$ with diameter $AB$ and center $O$. On the $(c)$ we take point $C$ such that the tangent at the $C$ intersects the line $AB$ at the point $E$. The perpendicular line from $C$ to $AB$ intersects the diameter $AB$ at the point $D$. On the $(c)$ we get the points $H,Z$ such that $CD = CH = CZ$. The line $HZ$ intersects the lines $CO,CD,AB$ at the points $S, I, K$ respectively and the parallel line from $I$ to the line $AB$ intersects the lines $CO,CK$ at the points $L,M$ respectively. We consider the circumcircle $(k)$ of the triangle $LMD$, which intersects again the lines $AB, CK$ at the points $P, U$ respectively. Let $(e_1), (e_2), (e_3)$ be the tangents of the $(k)$ at the points $L, M, P$ respectively and $R = (e_1) \cap (e_2)$, $X = (e_2) \cap (e_3)$, $T = (e_1) \cap (e_3)$. Prove that if $Q$ is the center of $(k)$, then the lines $RD, TU, XS$ pass through the same point, which lies in the line $IQ$.

Consider a semicircle with center $M$ and diameter $AB$. Let $P$ be a point in the semicircle, different from $A$ and $B$, and let $Q$ be the midpoint of the arc $AP$. The line parallel to $QP$ through $M$ intersects $PB$ at the point $S$. Prove that the triangle $PMS$ is isosceles.

A semicircle is erected over the segment $AB$ with center $M$. Let $P$ be one point different from $A$ and $B$ on the semicircle and $Q$ the midpoint of the arc of the circle $AP$. The point of intersection of the straight line $BP$ with the parallel to $P Q$ through $M$ is $S$. Prove that $PM = PS$ holds. [i](Karl Czakler)[/i]

Let $ABC$ be a triangle with $AB = \frac{AC}{2 }+ BC$. Consider the two semicircles outside the triangle with diameters $AB$ and $BC$. Let $X$ be the orthogonal projection of $A$ onto the common tangent line of those semicircles. Find $\angle CAX$.

(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 $\angle VAU$.