concurrency wanted, AE = AB, <IBG = < CBA (Kharkiv City IX 2014 - Ukr)

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concurrency wanted, AE = AB, <IBG = < CBA (Kharkiv City IX 2014 - Ukr)

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#1 h Mar 14, 2020, 6:46 AM

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In the triangle $ABC$ with $AB <AC$ is satisfied, let point $I$ be the center of the inscribed circle of this triangle. Point $E$ is selected on the side $AC$ such that $AE = AB$ . Point $G$ lies on the straight line $EI$ such that $\angle IBG = \angle CBA$ , and points $E$ and $C$ lie on opposite sides of $I$ . Prove that the line $AI$ , the perpendicular on $AE$ at point $E$ , and the bisector of angle $\angle BGI$ intersect at one point.

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#2 h Mar 14, 2020, 2:16 PM

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Let the bisector of $\angle BGI$ and the perpendicular line intersect at point $R$ ,and point $AGBRE$ are concyclic

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