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
- 2 :1
- 5: 8
- 8 : 5
- 1: 2
The Correct Option is C
Solution and Explanation
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): \(8:5\)
Top Questions on Quadrilaterals
Complete the following activity to prove that the sum of squares of diagonals of a rhombus is equal to the sum of the squares of the sides.
Given: PQRS is a rhombus. Diagonals PR and SQ intersect each other at point T.
To prove: PS\(^2\) + SR\(^2\) + QR\(^2\) + PQ\(^2\) = PR\(^2\) + QS\(^2\)
Activity: Diagonals of a rhombus bisect each other.
In \(\triangle\)PQS, PT is the median and in \(\triangle\)QRS, RT is the median.
\(\therefore\) by Apollonius theorem,
\[\begin{aligned} PQ^2 + PS^2 &= \boxed{\phantom{X}} + 2QT^2 \quad \dots \text{(I)} \\ QR^2 + SR^2 &= \boxed{\phantom{X}} + 2QT^2 \quad \dots \text{(II)} \\ \text{Adding (I) and (II),} \quad PQ^2 + PS^2 + QR^2 + SR^2 &= 2(PT^2 + \boxed{\phantom{X}}) + 4QT^2 \\ &= 2(PT^2 + \boxed{\phantom{X}}) + 4QT^2 \quad (\text{RT = PT}) \\ &= 4PT^2 + 4QT^2 \\ &= (\boxed{\phantom{X}})^2 + (2QT)^2 \\ \therefore \quad PQ^2 + PS^2 + QR^2 + SR^2 &= PR^2 + \boxed{\phantom{X}} \\ \end{aligned}\]
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