Question

Two equal chords ab and cd of a circle with centre o intersect each other at
e. Prove that ad

Answer 1

From the figure there are two triangles ΔAED and ΔCEB

To prove ΔAED and ΔCEB

From the figure we get,

<AED = <CEB (Vertically opposite angles are equal)

<A = < C   (since <A and <C are on the angle subtended on the same arc)

Similarly <D = <B

All angle are equal. Therefore ΔAED ≅ ΔCEB

To prove AD = CB

Since ΔAED ≅ ΔCEB

By CPCT (Corresponding parts of congruent triangles)

AE = CE, ED = EB and AD = CB

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Answer 2

Step-by-step explanation:

Intersecting Chord Theorem:

When two chords intersect each other in a circle, the product of the segments of each chord are equal.

Here, the two chord AB and CD intersect at a point X. So, the chord AB has two segments AX and XB. Similarly, the chord CD has two segments CX and XD.

AX * XB = CX * XD

Given: AX = 8 cm AB = 14 cm CX - XD = 8 cm

So, XB = AB - AX = 6 cm and CX = 8 + XD

Using the theorem,

AX * XB = CX * XD

8 * 6 = (8 + XD) * XD

48=8∗XD+X

D

2

48=8∗XD+XD2

Let XD = x

x 2 +8x−48=0

x2+8x−48=0

x2 +12x−4x−48=0

x2+12x−4x−48=0

x(x+12)−4(x+12)=0

x(x+12)−4(x+12)=0

(x+12)(x-4) =0

x = -12 or x =4

Since x represents the length of a segment, it can't be negative. Rejecting x =-12

So x = 4

XD = 4 cm

CX = 8 + XD = 8 = 4 = 12 cm

Hence, CD = CX + XD = 12 + 4 = 16 cm

I hope it helps!

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