Geometry and tangent-chord theorem problem?

Question

According to the figure, CA is tangent to the circle, centre O, at A. ABT and POT are straight lines.

Question:

Given that BT is equal to the radius of the circle, prove that:

1. $\angle ABP = 3 \angle OBP$
2. $\angle POA = 3 \angle BOT$
3. $\angle OBT = 2 (\angle BTO + \angle CAB)$

My attempt:

• $\angle AOP = 2 \angle ABP. $
• $\angle OBP = \angle OPB.$
• $\angle CAB = \angle APO.$
• $\angle BOA = 2 \angle APB.$

Answer 1

a):

$BOT=2PBO$

$ABO=2BOT=4PBO$

$ABP=ABO-PBO=4PBO-PBI=3PBO$

b):

$ABP+\frac{1}{3}ABP=2BOT$

$\frac{4}{3}ABP=2BOT$

$ABP=\frac{1}{2}POA$

$\frac{4}{3}(\frac{1}{2}POA)=2BOT$

Therefore:

$POA=3BOT$

c):

$ OBT=AOB+OBA$

$AOB=2ABP=2CAB$

$OBA=2BTO$

Therefore:

$OBT 2(CAB+BTO)$

Answer 2

1. $$\measuredangle ABP=\measuredangle OPB+\measuredangle OTB=\measuredangle OPB+\measuredangle TOB=$$ $$=\measuredangle OPB+\measuredangle OPB+\measuredangle OBP=3\measuredangle OBP.$$
2. $$\measuredangle POA=\measuredangle OTB+\measuredangle OAB=\measuredangle OTB+\measuredangle OBA=$$ $$=\measuredangle OTB+\measuredangle OTB+\measuredangle BOT=3\measuredangle BOT.$$
3. $$\measuredangle OBT=\measuredangle OAB+\measuredangle AOB=\measuredangle ABO+2\measuredangle CAB=$$ $$=\measuredangle BTO+\measuredangle BOT+2\measuredangle CAB=2(\measuredangle BTO+\measuredangle CAB).$$