(Geometry) Circle, angles and tangents problem Let P be an external point of a circle with center in O and also the intersection of two lines r and s that are tangent to the circle. If PAB is a triangle such that AB is also also tangent to the circle, find AÔB knowing that P = 40°.
I draw the problem:

Then I tried to solve it, found some relations, but don't know how to proceed.

I highly suspect that PAB is isosceles, but couldn't prove it.
 A: First of all, note that $\angle PAB + \angle PBA = 140^\circ$. That means that $\angle MAB + \angle NBA = 220^\circ$.
Then we see that $AO$ bisects $\angle MAB$, and $BO$ bisects $\angle NBA$, so $\angle OAB + \angle OBA = 110^\circ$.
Lastly, looking at the quadrilateral $AOBP$, we see that $x = 360^\circ - 40^\circ - 140^\circ - 110^\circ = 70^\circ$.
There is no reason to believe $\triangle PAB$ to be isosceles. In fact, from just the given information it might not be. If we move $A$ closer to $M$, we see that $AB$ touching the circle will force $B$ closer to $P$. It's just that you've happened to draw the figure symmetrically.
A: We know $\angle PON=70°$ and $\angle NOP=140°$.
And we also know that $OB$ and $OA$ are bisectors of $\angle NOT$ and $\angle TOM$ respectively. Therefore $$BOT+TOA=70°$$

A: Two tangents meet at a point. Therefore MP = NP. Only if OTP is a straight line, triangles MPN and APB are similar, hence both isoceles in this particular case - otherwise the triangles are not similar and only MPN is isoceles.
