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): $8:5$