Question 15.

In $\Delta ABC$, $AB=AC=12$ cm and D is a point on side BC such that $AD=8$ cm. If AD is extended to a point E such that $\angle ACB=\angle AEB$ then the length, in cm, of AE is

A
B
C
D
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Text Explanation
Based on the information provided, we can construct the following diagram;
image We have $AB = AC = 12$ cm, and $AD = 8$ cm. And $\angle ACB = \angle AEB$
Since the triangle is isosceles, $\angle ACB = \angle AEB = \angle ABC$ (or $\angle ABD$)
In $\triangle ABE$ and $\triangle ADB$,
$\angle BAE =\angle DAB$ and $\angle AEB = \angle ABD$. Therefore, by AA similarity criterion, $\triangle ABE \sim \triangle ADB$
Hence, we get,
$\dfrac{AB}{AD}=\dfrac{AE}{AB}$  or  $\dfrac{12}{8} = \dfrac{AE}{12}$  or  $AE=18$ cm.
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