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**niharika_kumar****Member**- From: Numeraland
- Registered: 2013-02-12
- Posts: 1,062

pls. help me out ...

Q.A circle is touching the side BC of a ΔABC at P and touching AB and AC produced at Q and R respt. Prove that AQ=1/2 (perimeter of ΔABC).

friendship is tan 90°.

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Let the centre of the circle be O.

Then we have OP = OQ (theyre radii of the circle) and ∠OPB =∠OQB = 90°. Hence BQ = BP.

Similarly CR = CP.

Therefore AB + BP = AQ = AR = AC + CP.

∴ AQ = AC + CP = ½(AC + CP + AC +CP) = ½(AB + BP + CP + AC) = ½ × perimeter.

*Last edited by Nehushtan (2013-10-25 07:30:31)*

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**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,108

hi niharika_kumar

The distances from a point outside a circle to the points where the tangent touches the circle are equal.

So for example AQ = AR and BQ = BP

I have called some distances w, x, y and z. Look at my diagram below for which.

By using the tangent rule above you will be able to do this quickly.

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**niharika_kumar****Member**- From: Numeraland
- Registered: 2013-02-12
- Posts: 1,062

thnks bob.

i was not able to construct the diagram.

nw its easy to do it

thnks again.

friendship is tan 90°.

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**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,108

I had three goes before getting a sensible diagram.

"AB produced at Q" was the clue to getting it correct. In all my previous attempts I had the points in this order A Q B along the line.

Then I realised the questioner was saying A B Q and so I had to twist A around to the other side of BC.

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**niharika_kumar****Member**- From: Numeraland
- Registered: 2013-02-12
- Posts: 1,062

thnx 2 nehushtan too fr giving the solution.

@bob i was confused as everytime i was gettin' some haphazard diagrams .

your diagram helped alot.

thnx

friendship is tan 90°.

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