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Question
In the given figure, two circles touch each other externally at point P. AB is the direct common tangent of these circles. Prove that :
- tangent at point P bisects AB,
- angles APB = 90°.
Solution
Draw TPT' as common tangent to the circles.
i. TA and TP are the tangents to the circle with centre O.
Therefore, TA = TP ...(i)
Similarly, TP = TB ...(ii)
From (i) and (ii)
TA = TB
Therefore, TPT' is the bisector of AB.
ii. Now in ΔATP,
∴ ∠TAP = ∠TPA
Similarly in ΔBTP, ∠TBP = ∠TPB
Adding,
∠TAP +∠TBP = ∠APB
But
∴ TAP + ∠TBP + ∠APB = 180°
`=>` ∠APB = ∠TAP + ∠TBP = 90°
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Two circles with centres O and O' touch each other at point L. Prove that, a tangent through L bisects the common tangent AB of the two circles at point M.
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To prove: M is a mid-point of the tangent AB or MA = MB.
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