Circle and tangents and chords

Geometry Level 3

A chord $AB$ lies on a circle with centre $O$ and radius $\text{3 cm}$ .

Chord $AB$ is in a circle with centre $O$ and radius $\text{3 cm}$ . Two tangents at point $A$ and point $B$ intersect at point $D$ . Another tangent parallel to $AB$ intersects $AD$ at $E$ and $BD$ at $F$ . Line $OD$ intersects the circle at $C$ .

If $AB=4.8 \text{ cm}$ , what is the length of $EF$ in $\text{cm}$ ?

The answer is 3.

2 solutions

Chris Lewis
Mar 13, 2020

We have $\sin \angle COA = \frac{\frac12 AB}{r} = \frac45$ .

Now note that points $OAEC$ form a kite; so $\angle COE = \frac12 \angle COA$ . Using standard identities, we have $\tan \angle COE = \frac12$ .

Finally, $EF=2CE=2\cdot 3\tan\angle COE = \boxed3$ .

A Former Brilliant Member
Mar 13, 2020

Let $\overline {AB}$ intersects $\overline {OC}$ at $G$ . Then $|\overline {OG}|=\sqrt {3^2-2.4^2}=1.8, |\overline {GC}|=3-1.8=1.2$ . $\triangle {OAG}$ and $\triangle {OAD}$ are similar. So $\dfrac{|\overline {OG}|}{|\overline {OA}|}=\dfrac{|\overline {OA}|}{|\overline {OD}|}$ . Hence $|\overline {OD}|=5$ and $|\overline {CD}|=5-3=2$ . Finally $\triangle {OEF}$ is similar to $\triangle {OAB}$ . So $\dfrac{|\overline {CD}|}{|\overline {EF}|}=\dfrac{|\overline {GD}|}{|\overline {AB}|}\implies |\overline {EF}|=4.8\times \dfrac{2}{2+1.2}=\boxed 3$ .

Circle and tangents and chords

The answer is 3.

2 solutions

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