Question

A circle has two equal chords AB and AC. Chord AD cuts BC in E. If $AC=12cm$ and $AE=8cm$, then AD is equal to

• A. 27
• B. 24
• C. 21
• D. 18

Answer 1

The correct option is D 18

In triangles ACD and ACE
A is common to both and $\angle D=\angle C$
(because $\angle D=\angle B$ and $\angle B=\angle C$)
$\therefore$ Triangles ACD and ACE are equiangular
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$\frac{AC}{AE}=\frac{AD}{AC}$
or, $\frac{12}{8}=\frac{AD}{12}$
or, $8\times AD=12\times 12$
or, $AD=\frac{12\times 12}{8}=18cm$