Found problems: 351
Show that an arbitrary quadrilateral can be divided into nine isosceles triangles.
$\vartriangle ABC$ is a triangle, $M$ the midpoint of $BC, D$ the projection of $M$ on $AC$ and $E$ the midppoint of $MD$. Prove that the lines $AE,BD$ are orthogonal if and only if $AB = AC$.
Consider the isosceles right triangle $ABC$, with $\angle A = 90^o$ and the point $M \in (BC)$ such that $\angle AMB = 75^o$. On the inner bisector of the angle $MAC$ take a point $F$ such that $BF = AB$. Prove that:
a) the lines $AM$ and $BF$ are perpendicular;
b) the triangle $CFM$ is isosceles.
The center $O$ of the circle $\omega$ passing through the vertex $C$ of the isosceles triangle $ABC$ ($AB = AC$) is the interior point of the triangle $ABC$. This circle intersects segments $BC$ and $AC$ at points $D \ne C$ and $E \ne C$, respectively, and the circumscribed circle $\Omega$ of the triangle $AEO$ at the point $F \ne E$. Prove that the center of the circumcircle of the triangle $BDF$ lies on the circle $\Omega$.
In the figure, $ABC$ and $DAE$ are isosceles triangles ($AB = AC = AD = DE$) and the angles $BAC$ and $ADE$ have measures $36^o$.
a) Using geometric properties, calculate the measure of angle $\angle EDC$.
b) Knowing that $BC = 2$, calculate the length of segment $DC$.
c) Calculate the length of segment $AC$ .
[img]https://1.bp.blogspot.com/-mv43_pSjBxE/XqBMTfNlRKI/AAAAAAAAL2c/5ILlM0n7A2IQleu9T4yNmIY_1DtrxzsJgCK4BGAYYCw/s400/2004%2Bobm%2Bl2.png[/img]
Let $ABC$ be an isosceles triangle with $AB = AC$. The points $D, E$ and $F$ are taken on the sides $BC, CA$ and $AB$, respectively, so that $\angle F DE = \angle ABC$ and $FE$ is not parallel to $BC$. Prove that the line $BC$ is tangent to the circumcircle of $\vartriangle DEF$ if and only if $D$ is the midpoint of the side $BC$.
Show that if a triangle has two excircles of the same size, then the triangle is isosceles.
(Note: The excircle $ABC$ to the side $ a$ touches the extensions of the sides $AB$ and $AC$ and the side $BC$.)
In an right isosceles triangle $ABC$, there are two points on the hypotenuse $AB, K$ and $M$, respectively, such that $KCM$ angle is $45^o$ (point $K$ lies between $A$ and $M$). Prove that $AK^2 + MB^2 = KM^2$
[img]https://cdn.artofproblemsolving.com/attachments/2/c/e7c57e0651e5a4c492cc4ae4b115bf68a7a833.png[/img]
Let $ABC$ be a triangle and $M$ be the midpoint of $AB$. If it is known that $\angle CAM + \angle MCB = 90^o$, show that triangle $ABC$ is isosceles or right.
Anton has an isosceles right triangle, which he wants to cut into $9$ triangular parts in the way shown in the picture. What is the largest number of the resulting $9$ parts that can be equilateral triangles?
A more formal description of partitioning. Let triangle $ABC$ be given. We choose two points on its sides so that they go in the order $AC_1C_2BA_1A_2CB_1B_2$, and no two coincide. In addition, the segments $C_1A_2$, $A_1B_2$ and $B_1C_2$ must intersect at one point. Then the partition is given by segments $C_1A_2$, $A_1B_2$, $B_1C_2$, $A_1C_2$, $B_1A_2$ and $C_1B_2$.
[img]https://cdn.artofproblemsolving.com/attachments/0/5/5dd914b987983216342e23460954d46755d351.png[/img]
An isosceles triangle $ABC$ ($AB = AC$) with an incircle of radius $r$ is given. We know that the point $M$ of the intersection of the medians of the triangle $ABC$ lies on this circle. Find the distance from the vertex $A$ to the point of intersection of the bisectrix of the triangle $ABC$.
(Grigory Filippovsky)
In a right triangle rectangle $ABC$ such that $AB = AC$, $M$ is the midpoint of $BC$. Let $P$ be a point on the perpendicular bisector of $AC$, lying in the semi-plane determined by $BC$ that does not contain $A$. Lines $CP$ and $AM$ intersect at $Q$. Calculate the angles that form the lines $AP$ and $BQ$.
Triangle $ABC$ is inscribed in circle with center $O$. Let $P$ be a point on arc $AB$ which does not contain point $C$. Perpendicular from point $P$ on line $BO$ intersects side $AB$ in point $S$, and side $BC$ in $T$. Perpendicular from point $P$ on line $AO$ intersects side $AB$ in point $Q$, and side $AC$ in $R$.
(i) Prove that triangle $PQS$ is isosceles
(ii) Prove that $\frac{PQ}{QR}=\frac{ST}{PQ}$
Restore the isosceles triangle $ABC$ ($AB = AC$) if the common points $I, M, H$ of bisectors, medians and altitudes respectively are given.
Given a triangle $ABC$. We construct two angle bisectors in the corners $A$ and $B$. Than we construct two lines parallel to those ones through the point $C$. $D$ and $E$ are intersections of those lines with the bisectors. It happens, that $(DE)$ line is parallel to $(AB)$. Prove that the triangle is isosceles.
In a trapezoid, the sum of the lengths of the side and the diagonal is equal to the sum of the lengths of the other side and the other diagonal. Prove that the trapezoid is isosceles.
In triangle $ABC$, medians $AA'$, $BB'$, $CC'$ are extended until they intersect with the circumcircle at points $A_0$, $B_0$, $C_0$, respectively. It is known that the intersection point M of the medians of triangle $ABC$ divides the segment $AA_0$ in half. Prove that the triangle $A_0B_0C_0$ is isosceles.
Let $ABC$ be a triangle and $A_1$ the foot of the internal bisector of angle $BAC$. Consider $d_A$ the perpendicular line from $A_1$ on $BC$. Define analogously the lines $d_B$ and $d_C$. Prove that lines $d_A, d_B$ and $d_C$ are concurrent if and only if triangle $ABC$ is isosceles.
In triangle $ABC$ we have $\angle C = 90^o$ and $AC = BC$. Furthermore $M$ is an interior pont in the triangle so that $MC = 1 , MA = 2$ and $MB =\sqrt2$. Determine $AB$
On the equal $ AC $ and $ BC $ of an isosceles right triangle $ ABC $ , points $ D $ and $ E $ are marked respectively, so that $ CD = CE $. Perpendiculars on the straight line $ AE $, passing through the points $ C $ and $ D $, intersect the side $ AB $ at the points $ P $ and $ Q $.Prove that $ BP = PQ $.
In a triangle $ABC$, right in $A$ and isosceles, let $D$ be a point on the side $AC$ ($A \ne D \ne C$) and $E$ be the point on the extension of $BA$ such that the triangle $ADE$ is isosceles. Let $P$ be the midpoint of segment $BD$, $R$ be the midpoint of the segment $CE$ and $Q$ the intersection point of $ED$ and $BC$. Prove that the quadrilateral $ARQP$ is a square.
As shown in the figure, given $\vartriangle ABC$ with $\angle B$, $\angle C$ acute angles, $AD \perp BC$, $DE \perp AC$, $M$ midpoint of $DE$, $AM \perp BE$. Prove that $\vartriangle ABC$ is isosceles.
[img]https://cdn.artofproblemsolving.com/attachments/a/8/f553c33557979f6f7b799935c3bde743edcc3c.png[/img]
The quadrilateral $ABCD$ is an isosceles trapezoid, where $AB\parallel CD$. The trapezoid is inscribed in a circle with radius $R$ and center on side $AB$. Point $E$ lies on the circumscribed circle and is such that $\angle DAE = 90^o$. Given that $\frac{AE}{AB}=\frac34$, calculate the length of the sides of the isosceles trapezoid.
Two isosceles triangles of the same area are located as shown in the figure. Find the angle $x$.
[img]https://cdn.artofproblemsolving.com/attachments/a/6/f7dbfd267274781b67a5f3d5a9036fb2905156.png[/img]
Let $ABC$ be a triangle. Let $D,E$ be the points outside of the triangle so that $AD=AB,AC=AE$ and $\angle DAB =\angle EAC =90^o$. Let $F$ be at the same side of the line $BC$ as $A$ such that $FB = FC$ and $\angle BFC=90^o$. Prove that the triangle $DEF$ is a right- isosceles triangle.