Q6 of 31 Page 218

Two circle drawn with centre A and B touch each other externally at C, O is a point on the tangent drawn at C, OD and OE are tangents drawn to the two circles of centers A and B respectively. If ∠ COD = 56°, ∠COE = 40, ∠ACD = x° and ∠BCE = y°. Let us prove that OD = OC = OE and y-x = 8

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∠ COD = 560


∠ COE = 400


∠ DAC and ∠ DOC are supplementary to each other since AD and AC are perpendicular to OD and OC respectively


⇒ ∠ DAC = (1800-560) = 1240


Since DA = AC so Δ DAC is isosceles


∠ ACD = 280


∠ COE and ∠ CBE are supplementary to each other since BC and BE are perpendicular to OC and OE respectively


⇒ ∠ CBE = (1800-400) = 1400


Since BC = BE so Δ CBE is isosceles


∠ BCE = 200


y-x = ∠ ACD-∠ BCE = 80


∠ CDO = ∠ OCD = 620 (∠ ADO and ∠ ACO is equal to 900)


Hence OD = OC


∠ OEC = ∠ OCE = 700(∠ OCB and ∠ OEB is equal to 900)


Hence OE = OC


So combining the above two we can say


OD = OC = OE


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