question on properties of circle 3

ukorov · #1 $Old$ November 1st, 2009, 10:30 AM

Referring to the attached pic:
(1) prove triangle BTQ ~ triangle QTA
(2) find TQ
(3) find the values of the radii

Grandad · #2 $Old$ November 2nd, 2009, 09:55 AM

Hello ukorov

Quote:

Originally Posted by ukorov $View Post$

Referring to the attached pic:
(1) prove triangle BTQ ~ triangle QTA
(2) find TQ
(3) find the values of the radii

Produce $QO_2$ to meet the circle at $R$ . Then $\angle TBQ = \angle BRQ$ (alternate segment)

$\Rightarrow \angle TBQ = \tfrac12 \angle BO_2Q$ (angle at centre)

$\angle TAO_1 = \angle TBO_2 = 90^o$ (angle between tangent and radius)

$\Rightarrow AO_1 \parallel BO_2$ (corresponding angles equal)

$\Rightarrow \angle AO_1T = \angle BO_2Q$ (corresponding angles)

But $\angle AQT =\tfrac12\angle AO_1T$ (angle at centre)

$\Rightarrow \angle AQT = \tfrac12\angle BO_2Q =\angle TBQ$ (proved)

So in $\triangle$ 's $BTQ, QTA$ :

$\angle AQT = \angle TBQ$ (proved)

$\angle ATQ = \angle QTB$ (same angle)

$\Rightarrow \triangle BTQ \sim \triangle QTA$

Grandad

ukorov · #3 $Old$ November 2nd, 2009, 11:22 AM

Quote:

Originally Posted by Grandad $View Post$

Hello ukorovProduce....

alright that makes sense now.
for question (2), is it to first prove that triangle AQB is right-angled, use Pythagarus Th. to find AB, and the rules of similar triangle to find TA, TO1, QO1, QO2?

Grandad · #4 $Old$ November 2nd, 2009, 12:42 PM

Hello ukorov

Quote:

Originally Posted by ukorov $View Post$

alright that makes sense now.
for question (2), is it to first prove that triangle AQB is right-angled, use Pythagarus Th. to find AB, and the rules of similar triangle to find TA, TO1, QO1, QO2?

Yes. Note that $PA=PQ=PB$ (tangents from point to circle). So $Q$ lies on a circle that has $AB$ as a diameter. So $\angle AQB =$ ?

Grandad

ukorov · #5 $Old$ November 2nd, 2009, 08:24 PM

Quote:

Originally Posted by Grandad $View Post$

Hello ukorovYes. Note that $PA=PQ=PB$ (tangents from point to circle). So $Q$ lies on a circle that has $AB$ as a diameter. So $\angle AQB =$ ?

Grandad

i saw that but i used angle AQT = a and angle BQR = (180 - a)/2 instead.

with adj. angle on straight line, angle AQB = 90