Let $A$ and $B$ be two points on a circle, and let the tangents to the circle at $A$ and $B$ meet at $P$. Prove that $AP = BP$.

Hint: Consider a reflection in the line which passes through P and the centre of the circle.

Work: Let the center of the circle be $O$, Then $OA = OB$ (radius of a circle), $AP,BP$ tangent implies that $\angle OAP=\angle OBP = 90^o$

From here, normally I can just say that these two triangles are congruent and therefore $AP = BP$

But this is the beginning of a upper division geometry course and I can't use the congruent properties yet.

I know the angles and the lengths will be preserved after I take the reflection as the hint given. But I don't see clearly how I can continue the proof.


Cant you just say that angles $PAB$ and $PBA$ are both equals to the half of arc $AB$, so that triangle is isosceles triangle?

  • $\begingroup$ Do you mind giving some further explanation and formalize it? Thank you $\endgroup$
    – Rico
    Apr 18 '19 at 4:15
  • $\begingroup$ Using this property(brilliant.org/wiki/alternate-segment-theorem-2) we can conclude that $\angle PAB = \frac{1}{2} \angle AOB$, similarly $\angle PBA = \frac{1}{2} \angle AOB$. So as $\angle PAB = \angle PBA$, than out triangle is isosceles. $\endgroup$ Apr 18 '19 at 4:27
  • $\begingroup$ I don't think we are assumed this knowledge. I think we can only use the fact that the reflection preserves length and angles. $\endgroup$
    – Rico
    Apr 18 '19 at 4:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.