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**niharika_kumar****Member**- From: Numeraland
- Registered: 2013-02-12
- Posts: 1,058

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).

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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****Moderator**- Registered: 2010-06-20
- Posts: 7,186

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

Children are not defined by school ...........The Fonz

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: 1,058

thnks bob.

i was not able to construct the diagram.

nw its easy to do it

thnks again.

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 7,186

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

Children are not defined by school ...........The Fonz

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: 1,058

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

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