In the figure on the above, P is a point outside the circle, with centre O, PA and PB are two tangents drawn from P to touch the circle at A and B respectively. We can find that
i) AP = BP
ii) angle APO = angle BPO
iii) angle AOP = angle BOP

angle OAP = angle OBP = 90°
(tangent ⊥ radius - to know more, read the information from our previous posts.)
△AOP and △BOP are congruent (RHS Property)
AP = BP
angle APO = angle BPO and angle AOP = angle BOP

**This videos is taken from ace-learning.org

Now, why not... TRY IT YOURSELF!

Question 1:

BC and CD are tangents to the following circle at B and D respectively. The centre of the circle is O. Given that angle BAD= 42° , find the obtuse angle of angle BCD.

It is given that YW and ZW are tangents to a circle at Y and Z respectively. The centre of the circle is O. If angle YXZ = 46°, find angle YWZ,

Answer: 88°

Questions are taken from ace-learning.org
**Answers are provided for you to check your answers.

Done By: Leona Ye

Once again, i would like to thank these websites:
http://library.thinkquest.org/C0110248/geometry/circlete.htm &
http://www.ace-learning.com.sg/sys/subjects/Qns.php
for providing us with the information we need in this post. Thank You! ;)

~By the way, please pardon us for the bad quality of the video and photos! :)