A, B, C and D lie on the circle. The chords AC and BD intersect at X.

Expert intro: This problem uses circle angle properties and vertically opposite angles to match the angles of two triangles.

Detailed walkthrough

Let us compare $riangle ADX$ and $riangle BCX$.

Step 1: Identify one pair of equal angles

Since the chords $AC$ and $BD$ intersect at $X$, the angles $\angle AXD$ and $\angle BXC$ are vertically opposite angles.

So,

$\angle AXD = \angle BXC.$

Step 2: Use angles in the same segment

Because $A, B, C, D$ lie on the same circle:

• $\angle ADX$ and $\angle BCX$ stand on the same chord $AB$? More precisely, the angle at $D$ and the angle at $C$ can be matched by the equal angles subtended by the same chord in the circle.
• Likewise, $\angle DAX$ and $\angle CBX$ are equal because they subtend the same arc.

Thus,

$\angle ADX = \angle BCX$

and

$\angle DAX = \angle CBX.$

Step 3: Conclude similarity

We now have two pairs of equal angles, so by AA similarity,

$\triangle ADX \sim \triangle BCX.$

A clean exam-style statement

• $\angle AXD = \angle BXC$ — vertically opposite angles are equal.
• $\angle ADX = \angle BCX$ — angles in the same segment are equal.
• $\angle DAX = \angle CBX$ — angles in the same segment are equal.

Therefore, the triangles are similar.

💡 Pitfall guide

A common mistake is to say only "all angles are the same" without naming the angle theorem used. In a proof question, each angle equality should be justified clearly, and you only need two matching angles to prove similarity by AA.

🔄 Real-world variant

If the intersecting chords were replaced by two secants meeting inside the circle, the same AA method often still works: first find a pair of vertically opposite angles at the intersection, then use equal angles subtended by the same chord or arc to match a second pair.

🔍 Related terms

vertically opposite angles, angles in the same segment, AA similarity