Question

Given: PA is tangent to the circle with center O.
Prove: triangle APB is similar to triangle CPA.


A, B, and C are points on the circle. P is outside the circle.

Answer 1

to show that they are similar you could show that they have the same three angles

Answer 2

thanks.

Answer 3

ok - so you know how to solve this now?

Answer 4

Here's what I wrote. Not sure if this proves it though:

Statements Reasons
__
PA tangent to the circle
Given
___
AB is a diameter forming chord ACB
By construction
m ∠ CAB = m ∠ PAB – m ∠ PAC Angle add. post.
m ∠ PAB = 90° Radius drawn to pt. of tangency
is # to tan.
m ∠ APB ≅ m ∠ CPA Given by construction
AB/AC = PB/AP SAS
Then ∆ APB ∼ ∆ CPA SAS
Or AB/AC = PB/AP = AP/PC SSS
Then ∆ APB ∼ ∆ CPA SSS

Answer 5

you can use a simpler AAA proof for this

Answer 6

you have shown that PAB = 90
you can also show that ACB = 90 (angle subtended by a diameter)

Answer 7

then:
CAB + CAP = PAB = 90

but:
CAB + CBA = 90 (because ACB=90)

therefore:
CBA = CAP

Answer 8

and therefore 3rd angle of both triangles must also be equal

Answer 9

I see. So, my solution is also correct?

Answer 10

I haven't read through your solution in detail - do you want me to?

Answer 11

reading now...

Answer 12

I don't see how:
AB/AC = PB/AP SAS

follows from:
m ∠ APB ≅ m ∠ CPA Given by construction

Answer 13

So, what do I need to do to correct this?

Answer 14

you could just use the AAA proof I gave above. :)

Answer 15

ok, thank you so much. You've been a great help!

Answer 16

yw :)