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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:

enter image description here Then I tried to solve it, found some relations, but don't know how to proceed. enter image description here I highly suspect that PAB is isosceles, but couldn't prove it.

Otávio Rapôso
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3 Answers3

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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.

Arthur
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  • Why is it that AO and BO bisect the angles? – Otávio Rapôso Nov 11 '16 at 10:26
  • Oh right, is it because OAT and OAM are congruous? (side-angle-side) – Otávio Rapôso Nov 11 '16 at 10:28
  • @OtávioRapôso That's one way to put it, yes. They're both right triangles, with all the same side lengths, so their other angles must match up. (I was going to put up this grand symmetry argument involving contradiction, but yours is just as good.) – Arthur Nov 11 '16 at 10:29
  • Ok, nice answer. I will wait to see if anyone has other nice solutions. – Otávio Rapôso Nov 11 '16 at 10:32
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    @OtávioRapôso Once you've noted that those two triangles are congruent, you could also do a similar argument by using the pentagon $AMONB$, where $\angle M = \angle N = 90^\circ, \angle A + \angle B = 220^\circ$ and $\angle O = 2x$. You should still get the same result. – Arthur Nov 11 '16 at 10:35
  • $\angle BNM = \angle AMN = 70^\circ$ also shows by quadrilateral $AMNB$ that $\angle MAB + \angle NBA = 220^\circ$ – Otávio Rapôso Nov 11 '16 at 10:43
  • True. There are many ways to chase the angles around. The point is, they are all connected to one another, and we may exploit that to get $x$ in the end. – Arthur Nov 11 '16 at 10:44
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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 $$BOT+TOA=70°$$ enter image description here

Seyed
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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.

Ajmir
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