Prove that of all the quadrilaterals circuscribed around a given circle, the square has the smallest perimeter.

Quadrilaterals may be obtained from an octagon by cutting along its diagonals (in $8$ different ways) . Can it happen that among these $8$ quadrilaterals (a) four (b ) five possess an inscribed circle? (P. M . Sedrakyan , Yerevan)

You can inscribe a circle in the pentagon $ABCDE$. It is also known that $\angle ABC = \angle BAE = \angle CDE = 90^o$. Find the measure of the angle $ADB$.

A quadrilateral is circumscribed around a circle. Prove that the straight lines connecting neighboring tangent points either meet on the extension of a diagonal of the quadrilateral or are parallel to it.

Nine numbers are arranged in a square table: $a_1 \,\,\, a_2 \,\,\,a_3$ $b_1 \,\,\,b_2 \,\,\,b_3$ $c_1\,\,\, c_2 \,\,\,c_3$ . It is known that the six numbers obtained by summing the rows and columns of the table are equal: $a_1 + a_2 + a_3 = b_1 + b_2 + b_3 = c_1 + c_2 + c_3 = a_1 + b_1 + c_1 = a_2 + b_2 + c_2 = a_3 + b_3 + c_3$ . Prove that the sum of products of numbers in the rows is equal to the sum of products of numbers in the columns: $a_1 b_1 c_1 + a_2 b_2c_2 + a_3b_3c_3 = a_1a_2a_3 + b_1 b_2 b_3 + c_1 c_2c_3$ . (V Proizvolov)

From which two statements about the trapezoid follows the third: 1) the trapezoid is tangential, 2) the trapezoid is right, 3) its area is equal to the product of the bases?

The quadrangle $ABCD$ is circumscribed around the circle with the centre $O$. Prove that $$\angle AOB+ \angle COD=180^o. $$

A quadrilateral is circumscribed around a circle. Its diagonals intersect at the center of the circle. Prove that the quadrilateral is a rhombus.

$ABCD$ is inscribed in unit circle $\Gamma$. Let $\Omega_1$, $\Omega_2$ be the circumcircles of $\vartriangle ABD$, $\vartriangle CBD$ respectively. Circles $\Omega_1$, $\Omega_2$ are tangent to segment $BD$ at $M$,$N$ respectively. Line A$M$ intersects $\Gamma$, $\Omega_1$ again at points $X_1$,$X_2$ respectively (different from $A$, $M$). Let $\omega_1$ be the circle passing through $X_1$, $X_2$ and tangent to $\Omega_1$. Line $CN$ intersects $\Gamma$, $\Omega_2$ again at points $Y_1$, $Y_2$ respectively (different from $C$, $N$). Let $\omega_2$ be the circle passing through $Y_1$, $Y_2$ and tangent to $\Omega_2$. Circles $\Omega_1$,$\Omega_2$, $\omega_1$, $\omega_2$ have radii $R_1$, $R_2$, $r_1$, $r_2$ respectively. Prove that $$r_1+r_2-R_1-R_2=1.$$ [img]https://cdn.artofproblemsolving.com/attachments/1/5/70471f2419fadc4b2183f5fe74f0c7a2e69ed4.png[/img] [url=https://www.geogebra.org/m/vxx8ghww]geogebra file[/url]

Considers a quadrilateral $[ABCD]$ that has an inscribed circle and a circumscribed circle. The sides $[AD]$ and $[BC]$ are tangent to the circle inscribed at points $E$ and $F$, respectively. Prove that $AE \cdot F C = BF \cdot ED$. [img]https://1.bp.blogspot.com/-6o1fFTdZ69E/X4XMo98ndAI/AAAAAAAAMno/7FXiJnWzJgcfSn-qSRoEAFyE8VgxmeBjwCLcBGAsYHQ/s0/2005%2BPortugal%2Bp5.png[/img]

Given a rectangle $A_1A_2A_3A_4$. Four circles with $A_i$ as their centres have their radiuses $r_1, r_2, r_3, r_4$; and $r_1+r_3=r_2+r_4<d$, where d is a diagonal of the rectangle. Two pairs of the outer common tangents to {the first and the third} and {the second and the fourth} circumferences make a quadrangle. Prove that you can inscribe a circle into that quadrangle.

Which of the following four statements are true and which are false? a) If a polygon inscribed in a circle is equilateral, then it is also equiangular. b) If a polygon inscribed in a circle is equiangular, then it is also equilateral. c) If a polygon circumscribed to a circle is equilateral, then it is also equiangular. d) If a polygon circumscribed to a circle is equiangular, then it is also equilateral.

$ABCD$ is inscribed in unit circle $\Gamma$. Let $\Omega_1$, $\Omega_2$ be the circumcircles of $\vartriangle ABD$, $\vartriangle CBD$ respectively. Circles $\Omega_1$, $\Omega_2$ are tangent to segment $BD$ at $M$,$N$ respectively. Line A$M$ intersects $\Gamma$, $\Omega_1$ again at points $X_1$,$X_2$ respectively (different from $A$, $M$). Let $\omega_1$ be the circle passing through $X_1$, $X_2$ and tangent to $\Omega_1$. Line $CN$ intersects $\Gamma$, $\Omega_2$ again at points $Y_1$, $Y_2$ respectively (different from $C$, $N$). Let $\omega_2$ be the circle passing through $Y_1$, $Y_2$ and tangent to $\Omega_2$. Circles $\Omega_1$,$\Omega_2$, $\omega_1$, $\omega_2$ have radii $R_1$, $R_2$, $r_1$, $r_2$ respectively. Prove that $$r_1+r_2-R_1-R_2=1.$$ [img]https://cdn.artofproblemsolving.com/attachments/1/5/70471f2419fadc4b2183f5fe74f0c7a2e69ed4.png[/img] [url=https://www.geogebra.org/m/vxx8ghww]geogebra file[/url]

a) The triangle $ABC$ was turned around the centre of the circumscribed circle by the angle less than $180$ degrees and thus was obtained the triangle $A_1B_1C_1$. The corresponding segments $[AB]$ and $[A_1B_1]$ intersect in the point $C_2, [BC]$ and $[B_1C_1]$ -- $A_2, [AC]$ and $[A_1C_1]$ -- $B_2$. Prove that the triangle $A_2B_2C_2$ is similar to the triangle $ABC$. b) The quadrangle $ABCD$ was turned around the centre of the circumscribed circle by the angle less than $180$ degrees and thus was obtained the quadrangle $A_1B_1C_1D_1$. Prove that the points of intersection of the corresponding lines ( $(AB$) and $(A_1B_1), (BC)$ and $(B_1C_1), (CD)$ and $(C_1D_1), (DA)$ and $(D_1A_1)$ ) are the vertices of the parallelogram.

A circle is inscribed in an trapezoid. Show that the diagonals of the trapezoid intersect at a point on the diameter of the circle perpendicular to the two parallel sides.

Let $ABCD$ be an tangential trapezoid, $E$ is a point of its diagonals intersection, $r_1,r_2,r_3,r_4$ -- the radiuses of the circles inscribed in the triangles $ABE$, $BCE$, $CDE$, $DAE$ respectively. Prove that $$1/(r_1)+1/(r_3) = 1/(r_2)+1/(r_4).$$

Quadrilateral $ABCD$ is circumscribed around circle $k$. Gind the smallest possible value of $$\frac{AB + BC + CD + DA}{AC + BD}$$, as well as all quadrilaterals with the above property where it is reached.

Segment $AB$ intersects two equal circles, is parallel to the line joining their centres, and all the points of intersection of the segment and the circles lie between $A$ and $B$. From the point $A$ tangents to the circle nearest to $A$ are drawn, and from the point $B$ tangents to the circle nearest to $B$ are also drawn. It turns out that the quadrilateral formed by the four tangents extended contains both circles. Prove that a circle can be drawn so that it touches all four sides of the quadrilateral. (P Kozhevnikov)