CAT 2023 Quant was dominated by Algebra followed by Arithmetic. In Arithmetic, the questions were dominated by topics like **Speed-time-distance**, **Mixture and Alligations**. This year, there was a surprise. The questions from **Geometry** were relatively on the lower side as compared to the previous years. There were 8 TITA Qs this year. Overall this section was at a medium level of difficulty.

Question 18 : 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 : 2
- 5 : 8
- 8 : 5
- 2 : 1

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\( \angle D A C = \angle D B C \)

*(Angles subtended by the chord DC on the same side.)*

\( \angle A D B = \angle A C B \)

*(Angles subtended by the chord AB on the same side.)*

\( \angle A E D = \angle B E C \)

*(Vertically Opposite angles.)*

Therefore, \( \triangle A E D \sim \triangle B E C \)

\( \frac { A E } { B E } = \frac { E D } { E C } = \frac { A D } { B C } = \frac { 4 } { 5 } \)

\( \angle A B D = \angle A C D \)

*(Angles subtended by the chord AD on the same side.)*

\( \angle B A C = \angle B D C \)

*(Angles subtended by the chord BC on the same side.)*

\( \angle A E B = \angle D E C \)

*(Vertically Opposite angles.)*

Therefore, \( \triangle A E B \sim \triangle D E C \)

\( \frac { A E } { E D } = \frac { B E } { E C } = \frac { A B } { D C } = \frac { 2 } { 1 } \)

\( \frac { A E } { E D } \times \frac { E D } { E C } = \frac { 2 } { 1 } \times \frac { 4 } { 5 } = \frac { 8 } { 5 } \)

Therefore, AE : EC = 8 : 5

The question is **" 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 " **

Choice C is the correct answer.

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