Question

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

• A. 2 :1
• B. 5: 8
• C. 8 : 5
• D. 1: 2

Answer 1

The Correct Option is C

Given that \(ABCD\) is a cyclic quadrilateral.
As \(∠ADB = ∠ACB\) and \(∠DAC = ∠DBC\) (angles subtended by chords on the same arc), \(△AED \) is similar to \(△BEC\) by AA similarity.
Similarly, \(△AEB\) is similar to \(△DEC\) by AA similarity.
Given the ratios \(AB : CD = 2:1\) and \(BC: AD = 5:4\).
From the similarity of \(△AED \) and \(△BEC\), we have \( \frac {AE}{BE} = \frac {AD}{BC} = \frac 45 \).
From the similarity of \(△AEB\) and \(△DEC\), we have \( \frac {BE}{CE} = \frac {AB}{CD} = \frac 21 \).
Multiplying these two ratios gives \( \frac {AE}{CE} = \frac {AE}{BE} \times \frac {BE}{CE} = \frac 45 \times \frac 21 = \frac 85 \).

Therefore, the correct option is (C): \(8:5\).