# Proofs of the Theorems on Circumscribable Quadrilaterals

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• Aug 15th 2010, 08:21 PM
jherv05
Proofs of the Theorems on Circumscribable Quadrilaterals

Here are some of the THEOREMS that needs to be proven;

THEOREM (5): A point is on the angle bisector of the angle if and only if the points is equidistant from the sides of the angle, that is, if and only if the lengths of the perpendicular segments from the point to the sides of the angle are equal.

THEOREM (6): The four angle bisectors of a quadrilateral are concurrent if and only if the quadrilateral is circumscribable.

THEOREM (7): A quadrilateral is circumscribable if and only if the incircles of the two triangles formed by a diagonal are tangent to each other.

THEOREM (8): The four sides of a circumscribable quadrilateral intersect the incircles of the two triangles formed by a diagonal of the quadrilateral to form four points, all of which are on a circle taht is concentric with the quadrilateral's inscribed circle.

THANK YOU VERY MUCH AND I HOPE TO SEE THE PROOFS AS SOON AS POSSIBLE. THANK YOU!
• Aug 16th 2010, 01:41 AM
Vlasev
What have you tried so far?
• Aug 16th 2010, 03:55 AM
jherv05
I have tried THEOREM (5) but I have doubts on it.

"if and only if" confused me with my proofs. I really don't know which comes first.
• Aug 16th 2010, 04:46 AM
Vlasev
If and only if proofs are bidirectional, meaning - you must prove both directions of the implication. Lets say a theorem is like this: Statement A if and only if Statement B. To prove it you need to prove the following two implications

Statement A implies Statement B. (If A, then B)
Statement B implies Statement A. (If B, then A)

For theorem 5 you need to show that if a point is on the angle bisector, then it must be equidistant from the sides. The other thing you need to prove is that if it is equidistant from the sides, it must lie on the bisector. I will help you with the first part.

Assume that a point P is on the bisector of some angle. You have two cases. If the angle is 180 degrees, it is a straight line. The bisector splits this angle into two, so we get 90 degrees and 90 degrees. You can see that any point on the bisector is trivially equidistant from the sides of the angle. Now, assume the angle is smaller than 180 degrees. Now follow the following drawing:

http://img827.imageshack.us/img827/739/angle.jpg

So point P is on the bisector. The angle is chopped into two equal smaller angles (a). Then, we have put down the lines from the bisector to the sides of the angle. They hit the sides at 90 degrees. Hence the third angles in these two triangles are the same (b). There was a theorem that if you have 1 common side (AP) and both the angles (a and b) adjacent to that side correspond in the two triangles, then the triangles are the same. Since the triangles are the same, the sides BP and CP must be the same. Thus we have proved one side of the implication. Can you prove the other?
• Aug 16th 2010, 05:30 AM
jherv05
I did that also. Thank you very much. I am correct if I say that I have to prove also that the point P (based on your figure) is on the angle bisector? because it's vice-versa, right?

In proving the to perpendicular segments are equidistant to each other, I use the idea of Triangle Congruence Postulates and Pythagorean Theorem.
• Aug 16th 2010, 05:40 AM
Vlasev
Yes, you assume it is equidistant and then you need to prove it is on the bisector.

For your second statement, yes it's ASA (angle-side-angle) from this page Congruence (geometry) - Wikipedia, the free encyclopedia.
• Aug 16th 2010, 06:12 AM
Vlasev
In theorem 6 does that mean that the circle is inscribed in the quadrilateral. If so, the proof should be straight-forward using theorem 5. However, I think it is false if the circle is on the outside of the quadrilateral!
• Aug 16th 2010, 06:23 AM
jherv05
yes it is inscribed in the quadrilateral. And it is actually in connection with the theorem (5). But I need to show the details. Can you please help me??
• Aug 16th 2010, 06:34 AM
Vlasev
Here is a hint. You start by having a circle inside the quadrilateral. Now this tells you something about the center of the circle and some distances involved. Keep in mind thm 5. It should be straightforward once you draw it.
• Aug 16th 2010, 06:33 PM
jherv05
I get it now. Thanks. Theorem (7) is quite simple but I am confused with theorem (8). can you please help me understand theorem (8)?
• Aug 16th 2010, 06:49 PM
Vlasev
I'm afraid I don't get that part. Could you draw a picture?
• Aug 16th 2010, 10:55 PM
jherv05
concentric circle maybe on the point of tangency in each side of the quadrilateral