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CAT 2023 Slot 1 — QA Question 22

Basics of QuadrilateralsEasy

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

Answer & solution

  • A

    2  :1

  • B

    5 : 8

  • C

    1 : 2

  • 8 : 5

Solution

Easy

In a cyclic quadrilateral, the diagonals create two pairs of similar triangles, because angles subtended by the same chord (in the same segment) are equal. Each similar pair gives a ratio of segments; multiply the right two ratios so that the unwanted segment EDED cancels, leaving AE:CEAE:CE.

A B C D E

Given ratios: AB:CD=2:1AB:CD=2:1 and BC:AD=5:4BC:AD=5:4. Diagonals AC,BDAC,BD meet at EE. Equal "same-segment" angles: ∠DAC=∠DBC\angle DAC=\angle DBC, ∠ADB=∠ACB\angle ADB=\angle ACB, ∠CDB=∠CAB\angle CDB=\angle CAB, ∠ACD=∠ABD\angle ACD=\angle ABD.

1

First similar pair: △AED∼△BEC\triangle AED\sim\triangle BEC (from ∠DAC=∠DBC\angle DAC=\angle DBC and ∠ADB=∠ACB\angle ADB=\angle ACB). Matching sides give

EDEC=ADBC=45(1)\frac{ED}{EC}=\frac{AD}{BC}=\frac{4}{5} \quad(1)
2

Second similar pair: △AEB∼△DEC\triangle AEB\sim\triangle DEC (from ∠CDB=∠CAB\angle CDB=\angle CAB and ∠ACD=∠ABD\angle ACD=\angle ABD). Matching sides give

AEED=ABCD=21(2)\frac{AE}{ED}=\frac{AB}{CD}=\frac{2}{1} \quad(2)
3

Multiply (1)×(2)(1)\times(2) so that EDED cancels, isolating AE:ECAE:EC:

EDEC×AEED=AEEC=45×21=85\frac{ED}{EC}\times\frac{AE}{ED}=\frac{AE}{EC}=\frac{4}{5}\times\frac{2}{1}=\frac{8}{5}

AE:CE=8:5AE:CE=\mathbf{8:5}.

CAT 2023 Slot 1 QA Q22: A quadrilateral ABCD is inscribed in a circle such that AB : CD = 2 : 1 and BC : AD = 5 : 4. If AC and BD inte — Solution | TheCATExam