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Circle Geometry Question (1 Viewer)

norez

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Aug 16, 2009
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HSC
2011
I was re-doing my Prelim exam from last year, and I still can't do this question. I feel like such a failure haha. I can't see how to do it. So:

O is the centre of the circle. TP is a tangent touching the circle at P. A is a point on the circle such that angle AOT is a right angle. AP cuts OT at Q.



Show that TP= TQ.
 

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xV1P3R

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Jan 1, 2007
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2010
angle OAQ = angle OPQ (Equal angles are opposite equal sides...)

So we let this be P = angle OAQ = angle OPQ

angle OPT = angle OPQ + angle QPT
90 = P + angle QPT (Angle between tangent and radius...)
angle QPT = 90 - P

Looking at triangle AQO
90 + angle OAQ + angle AQO = 180
P + angle AQO = 90
angle AQO = 90 - P
angle TQP = 90 - P (Vertically opposite..)

therefore TQ = TP (Equal sides opposite...)
 

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