In Fig. 9.14, common tangents AB and CD to two circles intersect at E. Prove that AB = CD
Solution:
Given, AB and CD are common tangents to two circles.
The tangents intersect at E.
we have to prove that AB = CD
From the figure,
E is the external point
The tangents drawn through external point E are EA, EC, EB and ED.
We know that the tangents drawn through an external point to a circle are equal.
So, EA = EC ---------------- (1)
EB = ED ---------------------- (2)
Adding (1) and (2),
EA + EB = EC + ED
From the figure,
AB = EA + EB
CD = EC + ED
So, AB = CD
Therefore, it is proved that AB = CD.
✦ Try This: In the given figure, common tangents AB and CD to the two circles with centres O1 and O2 intersect at E. Prove that AB = CD.
Given, AB and CD are the common tangents drawn to both the circles.
AB and CD intersect at the point E
As the tangents drawn on circle from same point
EA = EC
EB = ED
So by adding both we get
EA + EB = EC + ED
So we get
AB = CD
Therefore, it is proved that AB = CD.
☛ Also Check: NCERT Solutions for Class 10 Maths Chapter 10
NCERT Exemplar Class 10 Maths Exercise 9.3 Problem 7
In Fig. 9.14, common tangents AB and CD to two circles intersect at E. Prove that AB = CD
Summary:
In Fig. 9.14, common tangents AB and CD to two circles intersect at E. it is proven that AB = CD
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