tangents to a circle

niharika_kumar · 2013-10-25 02:37:07

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).

Nehushtan · 2013-10-25 03:15:56

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)

Bob · 2013-10-25 03:57:13

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

niharika_kumar · 2013-10-25 16:46:58

thnks bob.
i was not able to construct the diagram.
nw its easy to do it
thnks again.

Bob · 2013-10-25 19:35:32

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

niharika_kumar · 2013-10-26 16:40:02

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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