Cyclic Quadrilateral — Shorter Diagonal
October 31, 2024 2025-10-31 13:51Cyclic Quadrilateral — Shorter Diagonal
Cyclic Quadrilateral \(ABCD\): Find the Shorter Diagonal
Given \(BC = CD = 3\), \(DA = 5\), and \(\angle CDA = 120^\circ\). Use opposite-angle property \((\angle ABC=60^\circ)\), Law of Cosines for \(AC\) and \(AB\), then Ptolemy to extract \(BD\). Final: \( \tfrac{39}{7} \).
🧭 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 \(AB\)?
