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Proof Problem, Given: Segment PA is tangent to the circle with center O

Mathematics
Tutor: None Selected Time limit: 1 Day

Nov 29th, 2014

To prove Triangle APB is similar to Triangle APC

First note that Angle PAB is a right angle because the tangent PA must be perpendicular to the diameter AB at the point of contact.

Also angle ACB is a right angle because it is an angle in a semi -circle, so Angle ACP =180-90  =90 degrees

(ACP is a straight angle)

Hence Considering Triangle APB and Triangle APC

Angle PAB = Angle ACP= 90 degrees as shown above

Angle APB =Angle APC (same angle)

Angle PBA = Angle CAP ( Supplementary to 2 equal angles in the triangles)

Hence since the two triangles are equangular, they must be similar.

Triangle APB is similar to Triangle APC


Nov 29th, 2014

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Nov 29th, 2014
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Nov 29th, 2014
Dec 5th, 2016
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