# Synthetic Geometry Group-Sualeh's Proposal (Pakistan Round 1)

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 $AF=CE$.

###### Geometry-Pakistan's First Round (for IMO 2016). Note by Sualeh Asif
5 years, 2 months ago

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Due to lack of time I have uploaded pics of my solution to problem 1.   - 5 years, 2 months ago

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

- 5 years, 2 months ago

absolutely!

- 5 years, 2 months ago

@Calvin Lin @Xuming Liang Here is my submission!

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

- 5 years, 2 months ago

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

- 5 years, 2 months ago

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

- 3 years, 7 months ago