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Geometry Difficulty 6.1 National olympiad Prove it Saudi Arabia

Given a triangle $ABC$ , the internal bisector of $\hat{A}$ intersects $BC$ at $D$ and the external bisector of $\hat{A}$ intersects $BC$ at $E$ . Let $C_1$ be the circumcircle of $ADE$ . $AC$ intersects $C_1$ at $F$ . Let $C_2$ be the circumcircle of $ABF$ and the tangent of $C_2$ at $A$ intersects $C_1$ at $G$ . Prove that $AF = AG$ .

Solution

Let $\widehat{ABC} = 2\beta$ , $\widehat{BAC} = 2\alpha$ , $\widehat{ACB} = 2\gamma$ . Thus, the law of sines on triangle $AEF$ gives us

$\frac{AE}{AF} = \frac{\sin(\alpha + 2\beta)}{\sin(\alpha - \beta + \gamma)}.$

On the other hand, using the law of sines on $ABD$ , $ADF$ and the fact that $\widehat{BAD} = 180^\circ - \widehat{DAF}$ , we get that

$\frac{AB}{AF} = \frac{BD \sin(\widehat{BDA})}{DF \sin(\widehat{FDA})} = \frac{BD \sin(\alpha + 2\beta)}{DF \sin(\alpha - \beta + \gamma)}.$

Combining these two identities, we obtain that $\frac{AE}{AB} = \frac{DF}{DB}$ . Then since we also have $\widehat{FDB} = \widehat{FAE} = \widehat{EAB}$ , triangle $EAB$ is similar to triangle $FDB$ . Thus we find that $\widehat{FBE} = 2\beta$ . Using this and the fact that $A, D, F, E, G$ are on the same circle, we are able to compute that $\widehat{GEA} = \alpha - \beta + \gamma$ . But this was also computed to be equal to $\widehat{AEF}$ . Then $\widehat{GFA} = \widehat{GEA} = \widehat{AEF} = \widehat{AGF}$ , so $AF = AG$ as desired.

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