Question
B1 is a point on the side AC of ∆ ABC and B1B is joined. line is drawn through A parallel to B1B meeting BC at A1 and another line is drawn through C parallel to B1B meeting AB produced at C1. Then
∆AA1C & ∆BB1C AA1 ‖ BB1 (BB1) / (AA1) = (B1C) / (AC) By ratio of Similarity in ∆ACC1 ~ ∆ABB1 = BB1 ‖ CC1 = (BB1) / (CC1) = (AB1) / (AC) = (BB1) / (CC1) = (AC – B1C) / (AC) = (BB1) / (CC1) = 1 – ((B1C) / (AC) from (1) (BB1) / (CC1) = 1 – (BB1) / (AA1) BB1[1/CC1 + 1 / AA1] = 1 = (1 / CC1) + (1 / AA1) = (1 / BB1)More Geometry Questions
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- Question 5
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- Question 9
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