Power of a point

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Maverick848

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In the proof of the power of a point theorem, you have to show that the two triangles in circle are similar. HOw do you do this?
 
Please pardon my ignorance, but is this the Theorem you have in mind?

Thank you.

Eliz.
 
Maverick848 said:
In the proof of the power of a point theorem, you have to show that the two triangles in circle are similar. HOw do you do this?
Click to expand...

You'll use the Angle-Angle similarity theorem. The triangles in question will either SHARE an angle (if the "point" is on or outside the circle), or they will contain a pair of vertical angles (if the "point" is inside the circle. That will give you one pair of congruent angles. Then, you'll have to look for an angle in each which is measured by the same arc of the circle....this will give you another pair of congruent angles. Once you have two pairs of congruent angles, you can state that the triangles are similar.

If you haven't done so, please check out the link provided by Eliz., which has nicely labeled diagrams which you might find helpful.
 
Inscribed Angle Theorem

Okay so your proof makes sense but it relies on the Inscribed Angle Theorem (If two angles are inscribed on the same chord and on the same side of the chord, then they are equal. ).
So sorry its probably just intuitive but could you show me how to prove this as well.
 
Re: Inscribed Angle Theorem

Maverick848 said:
Okay so your proof makes sense but it relies on the Inscribed Angle Theorem (If two angles are inscribed on the same chord and on the same side of the chord, then they are equal. ).
So sorry its probably just intuitive but could you show me how to prove this as well.
Click to expand...

For a proof that the measure of an inscribed angle is half the measure of its intercepted arc, look here:

http://mathforum.org/library/drmath/view/55073.html

Once you have this theorem, it is easy to show that two inscribed angles which intercept the same (or equal) arcs in a circle must have the same measure.
 
Okay I am not really sure how that helps because in that formula you have diameter but in the inscribed angle thing you just have cords.
 
Maverick848 said:
In the proof of the power of a point theorem, you have to show that the two triangles in circle are similar. HOw do you do this?
Click to expand...
That was your original question. Lets look at one diagram:
ppower16ui.gif

Angles that intercept the same arc are congruent!
Therefore \(\displaystyle \L
\angle BAD \equiv \angle BCD\quad ,\quad \angle ABC \equiv \angle ADC\).
The angles at P are vertical angles and therefore congruent.
Because these two triangles have all angles congruent, the triangles are similar.
The other possible two cases in the ‘power point’ theorem work the same way!
 
Maverick848 said:
Okay I am not really sure how that helps because in that formula you have diameter but in the inscribed angle thing you just have cords.
Click to expand...

Maverick848.....your comment tells me that you did not read through the entire discussion on the page to which I referred. If you had read it, you would have seen how to handle a situation where neither chord which forms the angle goes through the center of the circle. In that case, you draw a diameter from the vertex of the angle, and then either add (if the center is in the interior of the angle) or subtract (if the center is in the exterior of the angle) the measures of the two angles formed, each of which WILL have one side which is a diameter of the circle.

I think you're waiting for one of us to do the entire proof for you. I doubt if you're going to be happy with the results....
 
Okay I understand everything except why "Angles that intercept the same arc are congruent!" I just need help proving that last part.
 
Maverick848 said:
Okay I understand everything except why "Angles that intercept the same arc are congruent!" I just need help proving that last part.
Click to expand...

That is just a well-known theorem in geometry.
It is found in any standard textbook on geometry.
Surely it is in your text.
I do not believe that any of us here are prepared to give such a long proof on line.
So do look it up.
 
Sorry, I do not have a geometry textbook since I do not need this for school. There must be an internet site that proves the theorem as well, right?

Also, I tried your suggestion about drawing the diameter but I am still confused since the diagram at the website you showed me had a central angle and the POP case does not, even after you draw a diameter.
 
Maverick848 said:
Sorry, I do not have a geometry textbook since I do not need this for school.
Click to expand...
There is no library at that school?
 
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