What have you tried so far?
Please help me to have some proofs of the theorems related to 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!
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:
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?
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.
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.