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Thread: Proofs of the Theorems on Circumscribable Quadrilaterals

  1. #16
    Senior Member
    Jul 2010
    Observe the following illustration. We have the quadrilateral ABCD. The quadrilateral's incircle's center is pt. O. Then there are the two incircles of the triangles formed by the diagonal. P,Q,R and S are the points where the incircles touch the quadrilateral's sides. Now we need to prove that the sides PO, QO, RO and SO are of equal length. Once we do this, the theorem is proved.

    The red lines in the above image are the lines connecting the vertices to the center of the incircle. Since we have an incircle these lines bisect their respective angles of the quadrilateral (from the previous theorems). Since we want to show that the sides PO, QO, RO and SO are of equal length then the triangles should be the same two by two. It suffices to show that APO and ASO are congruent triangles and then by the same logic it will follow that PBO and QBO are congruent and also RCO and QCO are congruent! Well, we already have a side (AO) and one angle (angles PAO and SAO from bisector). It'll be simple to just prove that AP = AS.

    This is where the right hand side of the drawing comes in. XY and ZY are tangents to the circle with center E and they intersect at point Y. We have that angles EXY and EZY are 90 degrees each (why?). Since XE = ZE and by theorem 5, EY must be a bisector. Hence XYE = ZYE and thus XEY = ZEY. From SAS (sides XE, ZE and YE and angles XEY and ZEY) triangles XYE and ZYE are congruent. Hence XY = ZY.

    With this in hand, we immediately have that AS = AP since AD and AB are tangent to the incircle of triangle ABD. Similarly PBO = QBO and RCO = QCO. Hence the sides PO, QO, RO and SO are all the same. Which means they lie on the same circle with center O. Thus the theorem is proved.
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  2. #17
    Aug 2010
    San Antonio, Sibulan, Negros Oriental, Philippines
    thanks a lot Sir.
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