<BLC=2<BAC wanted, symmedian and circumcircle related

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<BLC=2<BAC wanted, symmedian and circumcircle related

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Champions Tournament

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Source: Champions Tournament (Ukraine) - Турнір чемпіонів - 2005 Seniors p2

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#1 h Sep 6, 2020, 6:15 AM

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Given a triangle $ABC$ , the line passing through the vertex $A$ symmetric to the median $AM$ wrt the line containing the bisector of the angle $\angle BAC$ intersects the circle circumscribed around the triangle $ABC$ at points $A$ and $K$ . Let $L$ be the midpoint of the segment $AK$ . Prove that $\angle BLC=2\angle BAC$ .

This post has been edited 1 time. Last edited by parmenides51, Sep 6, 2020, 11:31 PM
Reason: changed Juniors to Seniors

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menpo

#2 h Sep 6, 2020, 6:39 AM

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Jafet Baca`s pdf

Lemma: Let ABD be a triangle and $E$ the second point of intersection of the $B$ −symmedian with its
circumcircle. Then $M$ , the midpoint of the chord $BE$ , is the center of spiral similarity which maps $AB$ to $BD$ .

Proof: Line $BE$ is a symmedian of $\triangle{ABD}$ , so $AD$ is a symmedian of $\triangle{EAB}$ and $\triangle{EDB}$ , therefore,
$\angle{MAB} = \angle{DAE} = \angle{DBE} = \angle{DBM}; \angle{BDM} = \angle{ADE} = \angle{ABE} = \angle{ABM}$
whence, $\triangle{AMB}$ , $\triangle{AED}$ , $\triangle{BMD}$ are similiar in the same orientation. Evidently, this special center of spiral
similarity inherits the aforesaid uniqueness. The result follows.
Because $BE$ is also a symmedian of $\triangle{AED}$ , we can show along the same lines that $\triangle{AME}$ , $\triangle{ABD}$
and $\triangle{EMD}$ are pairwise directly similar; thus, $M$ takes $AE$ to $ED$ . In summary, we have deduced the
following results.

Conclusion is trivial.

Attachments:

Spiral Similarity.pdf (256kb)

This post has been edited 2 times. Last edited by menpo, Sep 6, 2020, 6:44 AM

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