Question:

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

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When you see a condition like $\angle ACB = \angle AEB$ involving two angles subtending the same chord, think \textbf{cyclic quadrilateral}. Once you know points are concyclic, the \textbf{intersecting chords theorem} and \textbf{Stewart's theorem} can quickly give products like $BD \cdot DC$ and lead to elegant solutions.

Updated On: Jan 3, 2026

$18$
$16$
$20$
$14$

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The Correct Option is A

Approach Solution - 1

To find the length of $ AE $, we need to analyze the given isosceles triangle $ \Delta ABC $ where $ AB = AC = 12 \text{ cm} $ and point $ D $ is on $ BC $ such that $ AD = 8 \text{ cm} $. $ AD $ is extended to $ E $ such that $ \angle ACB = \angle AEB $.

Let's follow these steps:

Since $ \angle ACB = \angle AEB $, by the property of cyclic quadrilaterals, points $ A, B, C, E $ must lie on a circle.
We know $ AB = AC $, thus $ \angle ABC = \angle ACB $. Therefore, $ \angle ABC = \angle AEB $ (since $ A, B, C, E $ are concyclic).
This setup implies that $ \triangle ABD \sim \triangle ABE $ by the Angle-Angle criterion (as they share angle $ \angle ABD $ and have the angle $ \angle AEB $ in common).
By similarity, we can write the proportion: \[ \frac{AD}{AB} = \frac{AB}{AE} \] With given values, $ AD = 8 \text{ cm} $ and $ AB = 12 \text{ cm} $: \[ \frac{8}{12} = \frac{12}{AE} \]
Solving this proportion for $ AE $: \[ \frac{2}{3} = \frac{12}{AE} \quad \Rightarrow \quad AE = 12 \times \frac{3}{2} = 18 \text{ cm} \]

Thus, the length of $ AE $ is $ 18 \text{ cm} $.

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Approach Solution -2

Step 1: Show that $A, B, C, E$ are concyclic. Given: \[ \angle ACB = \angle AEB. \] Both angles subtend the same chord $AB$. Therefore, points $A, B, C, E$ lie on the same circle (they are concyclic).
Step 2: Use intersecting chords at point $D$. Point $D$ lies on chord $BC$ and also on chord $AE$, since $AD$ is extended to meet the circle again at $E$. By the intersecting chords theorem (for two chords intersecting at an interior point $D$): \[ AD \cdot DE = BD \cdot DC. \] We are given: \[ AD = 8 \text{ cm}. \] So, \[ 8 \cdot DE = BD \cdot DC. \quad \cdots (1) \]
Step 3: Apply Stewart's theorem in $\triangle ABC$ with cevian $AD$. Let: \[ BD = m,\quad DC = n,\quad BC = a = m + n, \] and \[ AB = AC = 12,\quad AD = 8. \] Stewart's theorem for cevian $AD$ in $\triangle ABC$ states: \[ AB^2 \cdot DC + AC^2 \cdot BD = BC \cdot (AD^2 + BD \cdot DC). \] Substitute the values: \[ 12^2 \cdot n + 12^2 \cdot m = (m + n)\big(8^2 + mn\big). \] So, \[ 144(m + n) = (m + n)(64 + mn). \] Since $m + n = BC \neq 0$, we can cancel it: \[ 144 = 64 + mn \quad \Rightarrow \quad mn = BD \cdot DC = 80. \quad \cdots (2) \]
Step 4: Find $DE$ and then $AE$. From (1) and (2): \[ 8 \cdot DE = BD \cdot DC = 80 \quad \Rightarrow \quad DE = \frac{80}{8} = 10. \] Therefore, \[ AE = AD + DE = 8 + 10 = 18 \text{ cm}. \] \[ \boxed{18} \]