# Show that the triangle formed by two tangent lines of a circle is an isosceles triangle.

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?
• 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. Apr 18 '19 at 4:27