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

• A. \(18\)
• B. \(16\)
• C. \(20\)
• D. \(14\)

Hint:

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.

Answer 1

The Correct Option is A

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, \[ AE = AD + DE = 8 + 10 = 18 \text{ cm}. \] \[ \boxed{18} \]