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The angle between a tangent and the radius drawn at the point of contact is a right angle.
Assume line $\overline{AB}$ is tangent to the circle at C but $\angle OCB< 90\degree$.
- Without loss of generality is applied here, we assume $\angle OCB$ is acute.
Then we can make a triangle $\triangle OCM$ with M on line $\overline{AB}$ so that $\angle OCM = \angle OMC$
So $\angle OCM$ is isosceles, and so $\overline{OC}=\overline{OM}$
But $\overline{OC}$ is a radius, so $\overline{OM}$ is a radius, so M lies on the circle.
This is a contradiction!
Hence, $\angle OCB$ has to be $90\degree$.
The angle between a tangent and chord equals the angle in the alternate segment.
Prove that $\angle DAE = \angle DBA$
- $\overline{OA} = \overline{ OD}$ (radii)
- $\angle OAE = 90\degree$ (tangent/radius theorem)
- $\triangle AOD$ is isosceles
- $\angle OAD = \angle ODA$ (angle in isosceles $\triangle$)
- $\angle DOA = 180 - 2 \times \angle OAD$
- $\angle DBA = \frac{1}{2} \times \angle DOA$ (angle at centre is twice the angle at circumference)
- $\therefore \angle DBA = 90 \degree - \angle OAD$
- $\angle DAE = 90\degree - \angle OAD$
- $\therefore \angle DBA = \angle DAE$
The two tangents drawn from a point to a circle are of equal length.
Prove that $\overline{AP} = \overline{BP}$
- $\overline{OA} = \overline{OB}$ (radii)
- $\overline{OP} = \overline{OP}$ (same segment)
- $\angle PAC = \angle PBC = 90\degree$ (angle between tangent and radius)
- $\triangle PAC \cong \triangle PBC$ (RHS)
- $\overline{AP} = \overline{BP}$ (corresponding sides of congruent $\triangle$s)
