Question

The chords AB and CD are equal and intersected at the point X. Prove that AX=CX and BX=DX.

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Answer 1

Chord AB = chord CD
AX+XB=CX+XD. (A-X-B, C-X-D) .....(1)
AX*XB=CX*XD (theorem of chords intersecting inside the circle).....(2)
Therefore from (1) & (2)
AX=CX and BX=DX
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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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