Found problems: 58
In a triangle $ABC$ let $E$ be the midpoint of the side $AC$ and $F$ the midpoint of the side $BC$. Let $G$ be the foot of the perpendicular from $C$ to $ AB$. Show that $\vartriangle EFG$ is isosceles if and only if $\vartriangle ABC$ is isosceles.
Let $ABCD$ be a parallelogram with an acute angle at $A$. Let $G$ be a point on the line $AB$, distinct from $B$, such that $|CG| = |CB|$. Let $H$ be a point on the line $BC$, distinct from $B$, such that $|AB| =|AH|$. Prove that triangle $DGH$ is isosceles.
[asy]
unitsize(1.5 cm);
pair A, B, C, D, G, H;
A = (0,0);
B = (2,0);
D = (0.5,1.5);
C = B + D - A;
G = reflect(A,B)*(C) + C - B;
H = reflect(B,C)*(H) + A - B;
draw(H--A--D--C--G);
draw(interp(A,G,-0.1)--interp(A,G,1.1));
draw(interp(C,H,-0.1)--interp(C,H,1.1));
draw(D--G--H--cycle, dashed);
dot("$A$", A, SW);
dot("$B$", B, SE);
dot("$C$", C, E);
dot("$D$", D, NW);
dot("$G$", G, NE);
dot("$H$", H, SE);
[/asy]
On the side $AB$ of an isosceles triangle $AB$ ($AC=BC$) lie points $P$ and $Q$ such that $\angle PCQ \le \frac{1}{2} \angle ACB$. Prove that $PQ \le \frac{1}{2} AB$.
Let $P$ be a point inside the triangle $ABC$ such that $AB = BC$, $\angle ABC = 80^o$, $\angle PAC = 40^o$ and $\angle ACP = 30^o$. Find $\angle BPC$.
(G Galperin)
In the triangle $ABC, AC=BC, \angle C=90^o, D$ is the midpoint of $BC, E$ is the point on $AB$ such that $AD$ is perpendicular to $CE$. Prove that $AE=2EB$.
In $\vartriangle ABC, AB=AC=14 \sqrt2 , D$ is the midpoint of $CA$ and $E$ is the midpoint of $BD$. Suppose $\vartriangle CDE$ is similar to $\vartriangle ABC$. Find the length of $BD$.
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 $\angle ACB = 90^{\circ}$.
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 $\angle ACB = 90^{\circ}$.
Let $ABCD$ be a parallelogram with an acute angle at $A$. Let $G$ be a point on the line $AB$, distinct from $B$, such that $|CG| = |CB|$. Let $H$ be a point on the line $BC$, distinct from $B$, such that $|AB| =|AH|$. Prove that triangle $DGH$ is isosceles.
[asy]
unitsize(1.5 cm);
pair A, B, C, D, G, H;
A = (0,0);
B = (2,0);
D = (0.5,1.5);
C = B + D - A;
G = reflect(A,B)*(C) + C - B;
H = reflect(B,C)*(H) + A - B;
draw(H--A--D--C--G);
draw(interp(A,G,-0.1)--interp(A,G,1.1));
draw(interp(C,H,-0.1)--interp(C,H,1.1));
draw(D--G--H--cycle, dashed);
dot("$A$", A, SW);
dot("$B$", B, SE);
dot("$C$", C, E);
dot("$D$", D, NW);
dot("$G$", G, NE);
dot("$H$", H, SE);
[/asy]
Consider an isosceles triangle $ABC$, where $AB = AC$. $D$ is a point on the $AC$ side and $P$ a point on the segment $BD$ so that the angle $\angle APC = 90^o$ and $ \angle ABP = \angle BCP $. Determine the ratio $AD: DC$.
A circle $C$ with center $O$ on base $BC$ of an isosceles triangle $ABC$ is tangent to the equal sides $AB,AC$. If point $P$ on $AB$ and point $Q$ on $AC$ are selected such that $PB \times CQ = (\frac{BC}{2})^2$, prove that line segment $PQ$ is tangent to circle $C$, and prove the converse.
Let $ABC$ be a triangle with $\angle B > \angle C$. Let $P$ and $Q$ be two different points on line $AC$ such that $\angle PBA = \angle QBA = \angle ACB $ and $A$ is located between $P$ and $C$. Suppose that there exists an interior point $D$ of segment $BQ$ for which $PD=PB$. Let the ray $AD$ intersect the circle $ABC$ at $R \neq A$. Prove that $QB = QR$.
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]
Find all isosceles triangles that cannot be cut into three isosceles triangles with the same sides.
Show that an arbitrary quadrilateral can be divided into nine isosceles triangles.
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 $\angle ACB = 90^{\circ}$.
Restore the isosceles triangle $ABC$ ($AB = AC$) if the common points $I, M, H$ of bisectors, medians and altitudes respectively are given.
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.
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 $\angle ACB = 90^{\circ}$.
Each point on the plane is colored either red, green, or blue. Prove that there exists an isosceles triangle whose vertices all have the same color.
Consider an isosceles triangle $ABC$ with $AB = AC$. Let $\omega(XYZ)$ be the circumcircle of the triangle $XY Z$. The tangents to $\omega(ABC)$ through $B$ and $C$ meet at the point $D$. The point $F$ is marked on the arc $AB$ of $\omega(ABC)$ that does not contain $C$. Let $K$ be the point of intersection of lines $AF$ and $BD$ and $L$ the point of intersection of the lines $AB$ and $CF$. Let $T$ and $S$ be the centers of the circles $\omega(BLC)$ and $\omega(BLK)$, respectively. Suppose that the circles $\omega(BTS)$ and $\omega(CFK)$ are tangent to each other at the point $P$. Prove that $P$ belongs to the line $AB$.
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 $\angle ACB = 90^{\circ}$.
From the vertices of a regular 27-gon, seven are chosen arbitrarily. Prove that among these seven points there are three points that form an isosceles triangle or four points that form an isosceles trapezoid.
Let $ABC$ be a triangle with $\angle B > \angle C$. Let $P$ and $Q$ be two different points on line $AC$ such that $\angle PBA = \angle QBA = \angle ACB $ and $A$ is located between $P$ and $C$. Suppose that there exists an interior point $D$ of segment $BQ$ for which $PD=PB$. Let the ray $AD$ intersect the circle $ABC$ at $R \neq A$. Prove that $QB = QR$.
A triangle is cut into several (not less than two) triangles. One of them is isosceles (not equilateral), and all others are equilateral. Determine the angles of the original triangle.