Geometry Problem 2

$ABCD$ be a convex quadrilateral. Let $E = AB \cap CD$, $F = AD \cap BC$, $P = AC \cap BD$, and let $O$ the foot of the perpendicular from $P$ to the line $EF$. Prove that $\angle BOC = \angle AOD.$

Easy Math Editor

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Comments

Have u framed this question ? Because this question is somehow similar to the 2003 INMO P1.

Shrihari B - 5 years, 5 months ago

This is a China TST

Alan Yan - 5 years, 5 months ago

Sorry friend but i do not consider my level to be that of the China TST. Even India TST is far away. So could u pls post some solvable problems ? Some NMO level(solvable)

Shrihari B - 5 years, 5 months ago

@Shrihari B – Okay. I will post another. But in the meantime, here is my solution.

It suffices to prove $\angle COP = \angle AOP$ and $\angle BOP = \angle DOP$ . Extend $BD$ to intersect line $EF$ at $S$ . Extend $AC$ to intersect $EF$ at $T$ . We have that $(BPDS)$ and $(ETFS)$ are harmonic bundles. $(BPDS$ harmonic and $\angle POS = 90^{\circ}$ proves that $\angle BOP = \angle POD$ .

$E(BPDS)$ harmonic implies $E(APCT)$ harmonic. This information along with $\angle POT = 90^{\circ}$ proves $\angle COP = \angle AOP$ . Hence, we are done.

Alan Yan - 5 years, 4 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$