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eGurucool CBSE Class X: Math Query 1

By on Nov 14, 2005 in On A Whim | 3 comments

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I was given this question by my math teacher and I could not solve it, so I decided to put it up here. Have a go at it, it is based on concepts from ‘Similar Triangles’. The only hint she gave was that we have to draw altitudes and use the theorem ‘The ratio of the areas of two similar triangles is equal to the ratio of the squares of their corresponding sides’ and ‘The areas of two similar triangles are in the ratio of the squares of the corresponding sides’. I don’t know if this hint is true, so keep your mind open to other options.

The Question: (Refer to figure given at the beginning of the post) A triangle ABC is given. AD, BE and CF are ANY line segments intersecting the sides BC, AC and AB, respectively, at D, E and F. NOTHING ELSE is given about the nature of these line segments, i.e., they are not medians, altitudes, angles bisectors, etc.
Prove that (AP/PD) + (BP/PE) + (CP+PF) = 1.

I have no idea if the question is correct even. If you are an eGurucool personnel, then send back the answer using eGurucool. Otherwise, if you are a very kind mathematician surfing the web, answer the question in a comment or click here to email me.

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3 Comments

  1. Great Quizzard

    November 15, 2005

    Post a Reply

    Ah, forgot one thing. Th line segments AD, BE and CF intersect each other at point P, as shown in the figure.

  2. Great Quizzard

    November 19, 2005

    Post a Reply

    Ooooooooopppppppppsssssss! Another blunder! The ‘to prove’ for this question is something else. It should be
    (PD/AD) + (PE/BE) + (PF/CF) = 1.
    If you are an eGurucool guy then sorry, our teacher had given us the wrong question. She corrected it later but told us to think about the solution.

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