){capmDx"p"t $E  ("T.GMAnd, if this doesn't convince you, the cynic, then check out the proof below.ֱ6jL  ("T8VDrag either A or B along the semicircle and observe the measurements below as you do. u do. And, if this doesn't convice you, the cynic, then check out the proof below.*n5  ("T_ 7%56V  Proof: Suppose that a quadrilateral ABCD is inscribed in a semicircle of diameter x. Then, the measure of angle DAC = the measure of angle DBC = /2 since the measure of an angle whose vertex lies on a circle is equal to 1/2 the measure of its intercepted arc. Therefore, DAC and DBC are right triangles. Therefore, by the Pythagorean Theorem, a^2 + n^2 = x^2 and c^2 + m^2 = x^2. Recall the result of Ptolemy's Theorem. That is, given a convex quadralateral ABCD which is inscribed in a circle, then for sides a,b,c,x and diagonals n and m, ac + bx = mn. Next, square Ptolemy's result to get that m^2 n^2 = a^2 c^2 + 2abcx + b^2 x^2. Substitute the expressions for m^2 and n^2 from earlier so that (x^2 - c^2)(x^2 - c^2) = a^2 c^2 + 2abcx + b^2 x^2; x^4 -(a^2 + c^2)x^2 + a^2 c^2 = a^2 c^2 + 2abcx + b^2 x^2; x^4 -(a^2 + b^2 + c^2)x^2 - 2abcx = 0; x^3 -(a^2 + b^2 + c^2)x - 2abc = 0, as required. QED P _cPۀ < _cP۠ @ _cP HM ("TDHBCmCH ("TCHCCm GH  ("TxHCCmBCm?FjvH ("Tm1Ȑ2P  4p2P px = "p#($L@ڐ $ $' & 4XD"ߔ | 99999999wwwwkLength(Segment x) = "@;"."4"@(199"@wwwwww4L>$@!l"p"H"1|"Yp)" I""H"-$wwwww"wwwwww"wwwwwwww,8+$@!l"p"H"1|"Yp)" I""H"-$wwwww"wwwwww"wwwwwwww,8+$@!l"p"H"1|"Yp)" I""H"-$wwwww"wwwwww"wwwwwwww,8+$@!l"p"H"1|"Yp)" I""H"-$wwwww"wwwwww"wwwwwwww,8+