Cyclic Quadrilateral

Date: 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/