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Circle Midpoint Config
Fuyuki   0
24 minutes ago
In triangle ABC, point D is the midpoint of BC. Let the second intersection of AD and (ABC) be E. Then, F is the intersection of EC and AB. G is the intersection of BE and AC. Prove that BC is parallel to FG.
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Fuyuki
24 minutes ago
0 replies
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A weird problem
jayme   0
2 hours ago
Dear Mathlinkers,

1. ABC a triangle
2. 0 the circumcircle
3. I the incenter
4. 1 a circle passing througn B and C
5. X, Y the second points of intersection of 1 wrt BI, CI
6. 2 the circumcircle of the triangle XYI
7. M, N the symetrics of B, C wrt XY.

Question : if 2 is tangent to 0 then, 2 is tangent to MN.

Sincerely
Jean-Louis
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jayme
2 hours ago
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nice geo
Melid   3
N Apr 27, 2025 by Melid
Source: 2025 Japan Junior MO preliminary P9
Let $ABCD$ be a cyclic quadrilateral, which is $AB=7$ and $BC=6$. Let $E$ be a point on segment $CD$ so that $BE=9$. Line $BE$ and $AD$ intersect at $F$. Suppose that $A$, $D$, and $F$ lie in order. If $AF=11$ and $DF:DE=7:6$, find the length of segment $CD$.
3 replies
Melid
Apr 23, 2025
Melid
Apr 27, 2025
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nice geo
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Source: 2025 Japan Junior MO preliminary P9
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Melid
11 posts
Melid
#1 h  Apr 23, 2025, 3:01 PM • 1 Y
Y by Rounak_iitr
Let $ABCD$ be a cyclic quadrilateral, which is $AB=7$ and $BC=6$. Let $E$ be a point on segment $CD$ so that $BE=9$. Line $BE$ and $AD$ intersect at $F$. Suppose that $A$, $D$, and $F$ lie in order. If $AF=11$ and $DF:DE=7:6$, find the length of segment $CD$.
This post has been edited 1 time. Last edited by Melid, May 23, 2025, 2:43 PM
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Melid
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Melid
#2 h  Apr 23, 2025, 3:06 PM
Y by
sorry for my bad English...
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L_.
3 posts
L_.
#3 h  Apr 26, 2025, 6:44 PM
Y by 
As $ABCD$ is cyclic $\angle EDF = \angle ABC$, both supplementary to the same angle $\angle CDA$. Hence $ABC \sim EDF$ and $\angle AFB = \angle DFE= \angle EBC=\angle FBC$, so $AD \parallel AF \parallel BC$. Since every cyclic trapezoid is isosceles, it follows that $\boxed{\overline{CD}=7}$.
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Melid
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Melid
#4 h  Apr 27, 2025, 6:50 AM
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L_. wrote:
As $ABCD$ is cyclic $\angle EDF = \angle ABC$, both supplementary to the same angle $\angle CDA$. Hence $ABC \sim EDF$ and $\angle AFB = \angle DFE= \angle EBC=\angle FBC$, so $AD \parallel AF \parallel BC$. Since every cyclic trapezoid is isosceles, it follows that $\boxed{\overline{CD}=7}$.

Unfortunately, it is not ∠DFE=∠EBC...
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