PT and QT are two tangents to a circle with center O. prove that
angle POQ=2angle PTQ
I know PT=QT and both these line segments makes right angles with OP and OQ.
It isn't true (see an example in the attachment)! But you are right (I mean your remarks).
$\displaystyle TQOP$ is a cyclic quadrilateral (opposite angles are supplementary). (You can also prove this by using one of the right triangles you mentioned: $\displaystyle \frac{\alpha}2+\frac{\beta}2+90=180^\circ \rightarrow \alpha+\beta=180^\circ$.) So you know that $\displaystyle \alpha=2\beta$ if and only if $\displaystyle \alpha=120^\circ$ and $\displaystyle \beta = 60^\circ$.
first reply said it isn't true about the mentionned angles.without trying any calculations draw a circle and construct tangents from various outside points along an extended circle di ameter. you will see that the angles you mention change as you go from way out to way in. in fact if you make two radii intersect at 90 degrees you will form a square.
bj
make a drawing .you have two congruent triangles 90 deg angles atTPO andTQO .assign unity to the radius of the circle and amultiple of 1 to the length from 0 toT. angle POT equals cosine of 1/extended length and angle OTP is the complement of it these 1/2 half angles will be in the same ratio as the whole angles
extended length cos 1/el acute angle atcenter acute angle at T 1.15 .87 30 60
this is a special case.make your own chart . start at el 1.01 and take about 4 or 5 steps ending at 200. then define the range of each set of the acute angles. conclusion the ratio of the angles is a function of the distance of T from the center of the circle
bj