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Cyclic Quadrilateral Diagonal Ratio

CAT 2023 Slot 1 · Quantitative Ability · Medium · Geometry

This is a medium Quantitative Ability question from the CAT 2023 Slot 1 paper. It tests Geometry. The full answer key and a step-by-step explanation are below — try it yourself first, then reveal the solution.

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A quadrilateral ABCD is inscribed in a circle such that AB : CD = 2 : 1 and BC : AD = 5 : 4. If AC and BD intersect at the point E, then AE : CE equals

  1. A.

    1 : 2

  2. B.

    5 : 8

  3. C.

    8 : 5

  4. D.

    2 : 1

Answer

C

Explanation

For diagonals ACAC and BDBD intersecting at EE in a cyclic quadrilateral ABCDABCD: Area(ΔABD)Area(ΔCBD)=AECE\frac{\text{Area}(\Delta ABD)}{\text{Area}(\Delta CBD)} = \frac{AE}{CE}

Area(ΔABD)=12⋅AB⋅AD⋅sin⁡(A)\text{Area}(\Delta ABD) = \frac{1}{2} \cdot AB \cdot AD \cdot \sin(A) Area(ΔCBD)=12⋅BC⋅CD⋅sin⁡(C)\text{Area}(\Delta CBD) = \frac{1}{2} \cdot BC \cdot CD \cdot \sin(C)

Since opposite angles of a cyclic quadrilateral sum to 180∘180^\circ, sin⁡(A)=sin⁡(C)\sin(A) = \sin(C). AECE=AB⋅ADBC⋅CD=(ABCD)×(ADBC)=(21)×(45)=85\frac{AE}{CE} = \frac{AB \cdot AD}{BC \cdot CD} = \left(\frac{AB}{CD}\right) \times \left(\frac{AD}{BC}\right) = \left(\frac{2}{1}\right) \times \left(\frac{4}{5}\right) = \frac{8}{5}

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