Solve using Butterfly Theorem. Let $PT$ and $PB$ be two tangents to a circle, $AB$ the diameter through $B$, and $TH$ the perpendicular from $T$ to $AB$. Then prove that $AP$ bisects $TH$. 
 A: I don't see how should one use the Butterfly Theorem, but here's another rather simple solution. 
Construct the symmetrical point of $B$ with respect to $P$ and call it $D$. We'll prove that $A-T-D$ are collinear. First well prove that $\triangle DTB \sim \triangle BTA$. Note that $\angle DBT = \angle BAT$, by Tangent-Chord Theorem, therefore it's enough to prove that $\frac{TB}{DB} = \frac{TA}{AB}$, which can easily be done by the Sine Theorem:
$$\frac{TA}{DB} = \frac{TA}{2\cdot PB} = \frac{\sin(\angle BPT)}{2\sin(\angle TBP)}= \frac{\sin(\angle TOA)}{2\sin(\angle TOA)} = \frac{TA}{2\cdot TO} = \frac{TA}{2 \cdot BO} = \frac{TA}{AB}$$
Hence the claim that $\triangle DTB \sim \triangle BTA$ is proven. Now $\angle ATD = \angle ATB + \angle BTD = 2\angle ATB = \pi$. Hence $A-T-D$ are collinear. Now using the Intercept Theorem on the parallel line $TH$ and $DP$ we have:
$$\frac{TH}{TC} = \frac{DB}{DP} = 2$$
Therefore we can conclude that $C$ is a midpoint of $TH$.
A: I don't know how to solve this by butterfly theorem but this is a simpler way.

Construction: Draw the tangent through A cutting PT extended at S. 
Note that $PB // TH // SA$. Then, $\triangle PTC \sim \triangle PSA$ and $\triangle ACH \sim \triangle APB$.
Fact: BY tangent properties, PB = BT and ST = SA. 
Then, $\dfrac {CH}{PB} = \dfrac {AC}{AP} =\dfrac {ST}{SP} =\dfrac {SA}{SP} = \dfrac {CT}{PT}=\dfrac {CT}{PB}$.
Result follows.
