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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.

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

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  • $\begingroup$ 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

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