🧭 Problem

Cyclic quadrilateral \(ABCD\) has side lengths \(BC=CD=3\) and \(DA=5\) with \(\angle CDA=120^\circ\). What is the length of the shorter diagonal of \(ABCD\)?

1️⃣ Opposite angles in a cyclic quadrilateral

2️⃣ Law of Cosines on \(\triangle ACD\) → \(AC\)

Let \(AC = x\). In \(\triangle ACD\):

3️⃣ Law of Cosines on \(\triangle ABC\) → \(AB\)

Let \(AB = y\). With \(AC=7\), \(BC=3\), \(\angle ABC = 60^\circ\):

4️⃣ Ptolemy’s Theorem on \(ABCD\) → \(BD\)

\[ AB\cdot CD + AD\cdot BC = AC\cdot BD. \]

Since \(AC=7\) and \( \tfrac{39}{7} \lt 7\), the shorter diagonal is \( \boxed{\tfrac{39}{7}} \) → option (D).

📝 Quick Check

New data (same pattern): In cyclic \(ABCD\), let \(BC=CD=6\), \(DA=10\), and \(\angle CDA=120^\circ\). What is \(AB\)?