Cyclic QuadrilateralDate: 05/22/2000 at 06:24:26 From: Cherriblossom Subject: Cyclic quadrilateral For an isosceles trapezium ABCD with AB parallel to DC and AB < CD, prove that: 1) angle ADC = angle BCD 2) ABCD is a cyclic quadrilateral 3) the diagonals of ABCD are equal Please help me - I'm stuck. Thank you, Cherriblossom Date: 05/22/2000 at 07:06:35 From: Doctor Floor Subject: Re: Cyclic quadrilateral Hi Cherriblossom, thanks for writing. Let us consider the isosceles trapezium: C--------D \ / A----B Let us drop perpendicular altitudes from A and B to E and F respectively on CD. C-E----F-D \| |/ A----B Of course ABFE is a rectangle. By Pythagoras' theorem we find: DF^2 = BD^2 - BF^2 = AC^2 - AE^2 (AC = BD and AE = BF) = CE^2 so that CE = DF, and consequently CF = DE. Now we find, again using Pythagoras' theorem, that: BC^2 = CF^2 + BF^2 = DE^2 + AE^2 (DE = CF and AE = BF) = AD^2 and we proved that BC = AD, and the diagonals are equal (in length). Together with AC = BD this shows that triangles ADC and BCD are congruent (SSS), and thus angle ADC = angle BCD. Finally, we see that: angle ACD = angle CDB (e.g. from the congruence of ADC and BCD) = 90 deg - angle DBF = 180 deg - angle DBA (angle DBA = 90 deg + angle DBF) and from angle ACD + angle DBA = 180 degrees we can conclude that ABCD is a cyclic quadrilateral. If you need more help, just write back. Best regards, - Doctor Floor, The Math Forum http://mathforum.org/dr.math/ |
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