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Question
Prove that the tangents drawn at the end points of a chord of a circle make equal angles with the chord.
Solution
Let AB be a chord of a circle with centre O, and let AD and BD be the tangents at A and B respectively.
Suppose OD meets AB at C.
We have to prove that ∠DAC = ∠DBC.
We know that, the line segment joining the centre to the external point, bisects the angle between two tangents.
So, ∠ADC=∠BDC ...(i)
In △DCA and △DCB, we have
DA = DB [Tangents from an external point are equal]
∠ADC = ∠BDC [From (i)]
DC=DC [Common]
∴ ∆DCA ≅ ∆DCB [By SAS congruency rule]
⇒∠DAC = ∠DBC [BY C.P.CT]
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