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Geometrical Exploits
Lemma 1.Given two pairs of parallel lines - (XA,YB) and (ZA,WB), see FIG 1. Assume that distances between the lines in the two pairs are equal, i.e., dist(XA, YB) = dist(ZA, WB). Then1. <XAB = <ABW 2. OA = OB. Proof:Let us drop perpendiculars from A and B onto WB and XA, respectively. Two right triangles ABH and ABG are equal. Indeed, they share a hypotenuse AB and AH = BG by the conditions of the Lemma. Therefore, <GAB = <HBA. However, <XAB = 180 - <GAB and <ABW = 180 - <HBA. Which proves 1.Further, <XAZ = <YBW since their sides are parallel. Subtracting this from 1 we get <OAB = <OBA so that the triangle OAB is isosceles. This proves 2. Now referring to FIG 2. we have the following
Corollary 1.1. <EAB = <ABC = <BCD.2. <CDE = <DEA.
Corollary 2.EA = AB = BC = CD.Indeed, each of the above segments serves as a hypotenuse of a right triangle built in the manner of ABH and ABG on FIG 1.
Corollary 3.AD = BE.Indeed, by Lemma 1.2 AD = BD and BD = BE.
Lemma 2.Two quadrilaterals ABCD and BCDE are equal.Proof:Both quadrilaterals are trapeze. We sum up equality of their corresponding elements in the following table:
Q.E.D.
Corollary 4.1. CD = DE.2. <BCD = <CDE
Theorem.ABCDE on FIG 2 is a regular pentagon.
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