In a circle, chord AB and chord CD intersect at E such that AE : EB = 2 : 3 and CE : ED = 5 : 2. If AB =x and CD = y, which of the following is true?

Correct — D. When two chords cross inside a circle, the products of their parts are equal: AE × EB = CE × ED.

Split AB = x in 2 : 3: AE = 2x⁄5, EB = 3x⁄5
Multiply: AE × EB = 6x²⁄25
Split CD = y in 5 : 2: CE = 5y⁄7, ED = 2y⁄7
Multiply: CE × ED = 10y²⁄49

Set equal: 6x²⁄25 = 10y²⁄49
Rearrange: x²⁄y² = (10 × 25) ⁄ (49 × 6) = 250⁄294
Cancel 2: x²⁄y² = 125⁄147 → option (d)

Why the others are wrong

• (a)
  x²⁄y² = 15⁄16 — 15⁄16 = 0.9375, while the chord rule gives x²⁄y² = 125⁄147 ≈ 0.850. It does not satisfy 6x²⁄25 = 10y²⁄49.
• (b)
  x⁄y = 6⁄15 — x⁄y = 6⁄15 = 2⁄5 would make x²⁄y² = 4⁄25 = 0.16, far from 125⁄147. The true x⁄y is √(125⁄147) = 5√15⁄21 ≈ 0.92.
• (c)
  x⁄y = 10⁄9 — 10⁄9 is greater than 1, which would make AB the longer chord. The rule gives x²⁄y² = 125⁄147, below 1, so AB is the shorter one.

The intersecting chords theorem: if chords AB and CD meet at E inside a circle, then AE × EB = CE × ED.

It comes from similar triangles. △AEC ~ △DEB, because angles in the same segment are equal (∠CAB = ∠CDB) and the vertically opposite angles at E are equal.

When only ratios are given, name each chord's length (x and y), split it by its ratio, and the theorem becomes an equation between x² and y².

Key facts

• Intersecting chords AB and CD meeting at E satisfy AE × EB = CE × ED.
• From an outside point P, a tangent PA and a secant PBC satisfy PA² = PB × PC.
• A length split in the ratio m : n has parts m⁄(m + n) and n⁄(m + n) of the whole.

Common traps

• Multiplying the ratio numbers alone: 2 × 3 = 6 and 5 × 2 = 10 are not equal, because the parts are fifths of x and sevenths of y.
• Stopping at x⁄y and hunting for a whole-number match, when the ratio of the chords is irrational here.
• Inverting the fraction at the last step and writing 147⁄125.

No directly related past PYQ was found.