Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 5.6 AIME, harder Prove it

II. (25 points) As shown in Figure 2, $M$ is a point inside the circle $\odot O$ with diameter $AB$ , and the extensions of $AM$ and $BM$ intersect $\odot O$ at points $C$ and $D$ , respectively. A perpendicular line $MN$ is drawn from point $M$ to $AB$ at point $N$ . A tangent line is drawn from point $C$ to $\odot O$ and intersects $MN$ at point $E$ . Connect $DE$ . Prove: $DE$ is a tangent to $\odot O$ .

Solution

As shown in Figure 7, connect $O C$ , $O D$ , $O E$ , $B C$ , $O E$ , and let $O E$ intersect $\odot O$ at point $F$ . Clearly,
$\begin{array}{l} \angle O C E=\angle O N E=90^{\circ} \\ \Rightarrow O, N, C, E \text { are concyclic } \\ \Rightarrow \angle B O C=\angle C E N . \end{array}$

Thus, $\triangle O B C \backsim \triangle E M C$
$\Rightarrow \frac{O C}{B C}=\frac{E C}{M C} \text {. }$

Also, $\angle O C E=\angle B C M$ , so $\triangle O C E \backsim \triangle B C M$
$\begin{array}{l} \Rightarrow \angle C O E=\angle C B M \\ \Rightarrow \angle C O F=\angle D O F . \end{array}$

It is easy to prove that $\triangle O C E \cong \triangle O D E$
$\Rightarrow \angle O D E=\angle O C E=90^{\circ} \text {. }$

Therefore, $D E$ is the tangent to $\odot O$ .

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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.