Hello ukorov Originally Posted by
ukorov Referring to the attached pic:
(1) prove triangle BTQ ~ triangle QTA
(2) find TQ
(3) find the values of the radii
Produce $\displaystyle QO_2$ to meet the circle at $\displaystyle R$. Then $\displaystyle \angle TBQ = \angle BRQ$ (alternate segment)
$\displaystyle \Rightarrow \angle TBQ = \tfrac12 \angle BO_2Q$ (angle at centre)
$\displaystyle \angle TAO_1 = \angle TBO_2 = 90^o$ (angle between tangent and radius)
$\displaystyle \Rightarrow AO_1 \parallel BO_2$ (corresponding angles equal)
$\displaystyle \Rightarrow \angle AO_1T = \angle BO_2Q$ (corresponding angles)
But $\displaystyle \angle AQT =\tfrac12\angle AO_1T$ (angle at centre)
$\displaystyle \Rightarrow \angle AQT = \tfrac12\angle BO_2Q =\angle TBQ$ (proved)
So in $\displaystyle \triangle$'s $\displaystyle BTQ, QTA$: $\displaystyle \angle AQT = \angle TBQ$ (proved)
$\displaystyle \angle ATQ = \angle QTB$ (same angle)
$\displaystyle \Rightarrow \triangle BTQ \sim \triangle QTA$
Grandad