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

Approach: Read the angle condition as a concyclic (cyclic-quadrilateral) clue and finish with the Power of a Point theorem, no similar-triangle bookkeeping needed.

Step 1: Show $A, B, C, E$ are concyclic. $E$ is on ray $AD$ beyond $D$, on the same side as $C$ relative to $AB$. The condition $\angle AEB = \angle ACB$ means segment $AB$ subtends equal angles at $C$ and $E$ on the same side — so $C$ and $E$ lie on one circle through $A$ and $B$. Hence $A, B, C, E$ are concyclic.

Step 2: Identify the secant lines. Line $A$–$D$–$E$ and line $B$–$D$–$C$ both pass through the interior point $D$, and each meets the circle at two of these four points: one secant hits $A$ and $E$, the other hits $B$ and $C$.

Step 3: Power of the point $D$. For two chords crossing at $D$: \[ DA \cdot DE = DB \cdot DC. \] But I want $AE = AD + DE$, so I instead use the power relation about the whole chord through $A$: since $AB$ is a chord and $\angle ABD = \angle AEB$ (isosceles base angle equals the inscribed angle), $AB$ is tangent-like to the circle through $B$, giving the cleaner relation $AB^2 = AD\cdot AE$.

Step 4: Plug in. $AB^2 = AD \cdot AE \Rightarrow 144 = 8 \cdot AE$.

Step 5: Solve. \[ AE = \frac{144}{8} = 18\ \text{cm}. \] Answer: option (a), $18$ cm.