These problems are my first submissions to Xuming's Geometry group.

However they are for the purpose of sharing the Geometry problems from Pakistan's First Round (for 2016)

Q1.The tangents at $A,B$ to the circumcircle$\omega$ of Triangle $ABC$ meet at $T$. The line through $T||AC$ meets $BC$ at $D$. Prove that $AD=CD$.

Q2. In a right angled triangle $ABC, \hat{C}=90^{\circ}, CD\perp AB$ at $D$ and the angle bisector of $\hat{B}$ intersects $CD,AC$ at $O,E$ respectively. Through $O$ introduce $FG \parallel AB$ such that $FG$ intersects $AC,BC$ at $F,G$ respectively. Prove that $AF=CE$.

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## Comments

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TopNewestDue to lack of time I have uploaded pics of my solution to problem 1.

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Good job Nihar ! The crux move was to identify that TBDA is cyclic, by wishful thinking.

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absolutely!

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@Calvin Lin @Xuming Liang Here is my submission!

@Nihar Mahajan ,@Sharky Kesa I thought you would be interested!

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Ill add in the solutions when requested. (Not before a week though)

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1st one is just some simple angle chase using alternate segment theorem as proved in the solution by @Nihar Mahajan.

2nd one is also just trivial: Solution to II.

$\Delta DBC \sim \Delta CBA$

=> $\dfrac{DB}{CB} = \dfrac{CB}{AB}$

=> $\dfrac{DO}{OC} = \dfrac{CE}{AE}$ --- (Int. Angle Bisector Theorem)

=> $\dfrac{AF}{CF} = \dfrac{CE}{AE}$

=> $\dfrac{AF}{AC} = \dfrac{CE}{AC}$ ---- (By rule of Dividendo)

=> $AF = AC$

K.I.P.K.I.G

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