Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 5.6 AIME, harder Prove it

As shown in Figure 7, quadrilateral $ABCD$ is a circumscribed convex quadrilateral, with points of tangency $E, H, F, G$ on sides $AB, BC, CD,$ and $DA$ respectively. Then $AC, BD, EF,$ and $GH$ are concurrent.

This is known as Newton's Theorem.

Solution

Proof: Let $E F$ intersect $A C$ at point $P$ . Then
$\begin{array}{l} \frac{A P}{A E}=\frac{\sin \angle A E P}{\sin \angle A P E}=\frac{\sin \angle C F P}{\sin \angle C P F}=\frac{P C}{C F} \\ \Rightarrow \frac{A P}{P C}=\frac{A E}{C F} . \end{array}$

Similarly, let $G H$ intersect $A C$ at $P^{\prime}, \frac{A P^{\prime}}{P^{\prime} C}=\frac{A G}{C H}$ .
But $A E=A G, C F=C H$ , thus, $\frac{A P}{P C}=\frac{A P^{\prime}}{P^{\prime} C}$ .
Therefore, point $P$ coincides with $P^{\prime}$ , meaning $A C, E F, G H$ are concurrent, or in other words, $A C$ passes through the intersection of $E F$ and $G H$ .
Similarly, $B D$ also passes through the intersection of $E F$ and $G H$ .

Thus, $A C, B D, E F, G H$ are concurrent.

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.