The center of circle passing through the midpoints of sides of isosceles triangle ABCConnecting midpoints of...
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The center of circle passing through the midpoints of sides of isosceles triangle ABC
Connecting midpoints of sides of a triangleprove that center of circle lie on the other circleFind the approximate center of a circle passing through more than three pointsProve that the midpoint of a certain segment is also the center of the escribed circle of a triangle.Center of circle tangent to hypotenuse in isosceles right triangleHelp me find a side of triangle given 2 sides and one circleProve that lines passing through the midpoints of sides of a triangle and the midpoints of cevians are also concurrentProof that 4 points lie on a circle and that center of this circle lies on the circumcircle of $triangle ABC$For I the incenter in △ABC, if AB+IC=AC+IB, then △ABC is isosceles.
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$begingroup$
The center of the circle passing through the midpoints
of sides of isosceles triangle $ABC$ lies on the circumcircle
of triangle $ABC$. If largest angle of the triangle is $x$ and smallest is $y$. find $x-y$.
geometry euclidean-geometry triangles circles
edited 11 hours ago
J. W. Tanner
14.4k1 gold badge10 silver badges30 bronze badges
asked 16 hours ago
mavericmaveric
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$endgroup$
add a comment |
$begingroup$
The center of the circle passing through the midpoints
of sides of isosceles triangle $ABC$ lies on the circumcircle
of triangle $ABC$. If largest angle of the triangle is $x$ and smallest is $y$. find $x-y$.
geometry euclidean-geometry triangles circles
edited 11 hours ago
J. W. Tanner
14.4k1 gold badge10 silver badges30 bronze badges
asked 16 hours ago
mavericmaveric
1,04312 bronze badges
$endgroup$
add a comment |
$begingroup$
The center of the circle passing through the midpoints
of sides of isosceles triangle $ABC$ lies on the circumcircle
of triangle $ABC$. If largest angle of the triangle is $x$ and smallest is $y$. find $x-y$.
geometry euclidean-geometry triangles circles
edited 11 hours ago
J. W. Tanner
14.4k1 gold badge10 silver badges30 bronze badges
asked 16 hours ago
mavericmaveric
1,04312 bronze badges
$endgroup$
The center of the circle passing through the midpoints
of sides of isosceles triangle $ABC$ lies on the circumcircle
of triangle $ABC$. If largest angle of the triangle is $x$ and smallest is $y$. find $x-y$.
geometry euclidean-geometry triangles circles
geometry euclidean-geometry triangles circles
edited 11 hours ago
J. W. Tanner
14.4k1 gold badge10 silver badges30 bronze badges
asked 16 hours ago
mavericmaveric
1,04312 bronze badges
edited 11 hours ago
J. W. Tanner
14.4k1 gold badge10 silver badges30 bronze badges
asked 16 hours ago
mavericmaveric
1,04312 bronze badges
edited 11 hours ago
J. W. Tanner
14.4k1 gold badge10 silver badges30 bronze badges
edited 11 hours ago
J. W. Tanner
14.4k1 gold badge10 silver badges30 bronze badges
edited 11 hours ago
J. W. Tanner
14.4k1 gold badge10 silver badges30 bronze badges
14.4k1 gold badge10 silver badges30 bronze badges
asked 16 hours ago
mavericmaveric
1,04312 bronze badges
asked 16 hours ago
mavericmaveric
1,04312 bronze badges
asked 16 hours ago
mavericmaveric
1,04312 bronze badges
1,04312 bronze badges
add a comment |
add a comment |
1 Answer
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$begingroup$
The center of any circle with $D$ and $E$ on it must pass through the (potentially extended) bisector of $angle A$. For this center to be on the circumcircle of $triangle ABC$, the only possibility is for the center to be $A$ itself.
$AF=AD$ since they are both radii of the same circle. $AD=DB$ since $D$ is the midpoint of $overline{AB}$. $overline{AF}perp overline{BC}$, since $triangle ABC$ is isosceles. Therefore, since $AB=2AF$, $angle B=30^circ$. That makes $angle A=120^circ$, so the difference between them is $90^circ$.
answered 15 hours ago
Matthew DalyMatthew Daly
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$begingroup$
The center of any circle with $D$ and $E$ on it must pass through the (potentially extended) bisector of $angle A$. For this center to be on the circumcircle of $triangle ABC$, the only possibility is for the center to be $A$ itself.
$AF=AD$ since they are both radii of the same circle. $AD=DB$ since $D$ is the midpoint of $overline{AB}$. $overline{AF}perp overline{BC}$, since $triangle ABC$ is isosceles. Therefore, since $AB=2AF$, $angle B=30^circ$. That makes $angle A=120^circ$, so the difference between them is $90^circ$.
answered 15 hours ago
Matthew DalyMatthew Daly
2,9181 silver badge21 bronze badges
$endgroup$
add a comment |
$begingroup$
The center of any circle with $D$ and $E$ on it must pass through the (potentially extended) bisector of $angle A$. For this center to be on the circumcircle of $triangle ABC$, the only possibility is for the center to be $A$ itself.
$AF=AD$ since they are both radii of the same circle. $AD=DB$ since $D$ is the midpoint of $overline{AB}$. $overline{AF}perp overline{BC}$, since $triangle ABC$ is isosceles. Therefore, since $AB=2AF$, $angle B=30^circ$. That makes $angle A=120^circ$, so the difference between them is $90^circ$.
answered 15 hours ago
Matthew DalyMatthew Daly
2,9181 silver badge21 bronze badges
$endgroup$
add a comment |
$begingroup$
The center of any circle with $D$ and $E$ on it must pass through the (potentially extended) bisector of $angle A$. For this center to be on the circumcircle of $triangle ABC$, the only possibility is for the center to be $A$ itself.
$AF=AD$ since they are both radii of the same circle. $AD=DB$ since $D$ is the midpoint of $overline{AB}$. $overline{AF}perp overline{BC}$, since $triangle ABC$ is isosceles. Therefore, since $AB=2AF$, $angle B=30^circ$. That makes $angle A=120^circ$, so the difference between them is $90^circ$.
answered 15 hours ago
Matthew DalyMatthew Daly
2,9181 silver badge21 bronze badges
$endgroup$
The center of any circle with $D$ and $E$ on it must pass through the (potentially extended) bisector of $angle A$. For this center to be on the circumcircle of $triangle ABC$, the only possibility is for the center to be $A$ itself.
$AF=AD$ since they are both radii of the same circle. $AD=DB$ since $D$ is the midpoint of $overline{AB}$. $overline{AF}perp overline{BC}$, since $triangle ABC$ is isosceles. Therefore, since $AB=2AF$, $angle B=30^circ$. That makes $angle A=120^circ$, so the difference between them is $90^circ$.
answered 15 hours ago
Matthew DalyMatthew Daly
2,9181 silver badge21 bronze badges
edited 15 hours ago
answered 15 hours ago
Matthew DalyMatthew Daly
2,9181 silver badge21 bronze badges
answered 15 hours ago
Matthew DalyMatthew Daly
2,9181 silver badge21 bronze badges
answered 15 hours ago
Matthew DalyMatthew Daly
2,9181 silver badge21 bronze badges
2,9181 silver badge21 bronze badges
add a comment |
add a comment |
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