vasudev@asgb.UUCP (08/13/84)
Use the following result: A triangle inscribed in a semi-circle is a right angled triangle. The corollary to which is: If the apex of the triangle lies outside the semi-circle then it is an acute angled triangle. Solution: Let ABCD be the square. Let E be the mid-point of AB. Let F be the mid-point of BC. Let G be the mid-point of CD. Let H be the mid-point of DA. Construct the semi-circles AH, HD AB, DC CF, FB such that they all lie in the square. Let semi-circle AH intersect semi-circle AB at I. Let semi-circle HD intersect semi-circle DC at J. Join IJ. Construct I'J' such that I'J' < IJ and is completely within IJ. The triangles are: AI'B, DJ'C, AI'H, HI'D, HI'J' and BI'F, I'J'F and FJ'C. Since the apexes of all the triangles lie outside the semi-circle, they are all acute. -QED -asgb!vasudev