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.
\(∠ADB = ∠ACB  \)          (Angle subtended by chord on the same side of arc)
\(∠DAC = ∠DBC  \)          (Angle subtended by chord on the same side of arc
\(⇒ △AED \) and \(△BEC\) are similar triangles.
Similarly, \(△AEB\) and \(△DEC\) are also similar using AA similarity property.
Given that,
\(AB : CD = 2:1\)
and \(BC: AD = 5:4\)
\(\frac {AE}{BE} = \frac {AD}{BC} = \frac 45 \)         (Similar Triangles \(△AED\) and \(△BEC\))

\(\frac {BE}{CE} = \frac {AB}{CD} =\frac 21\)         (Similar Triangles \(△AEB\) and \(△DEC\))
On multiplying both,
\(\frac {AE}{CE} = \frac 85\)

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