Associated Topics || Dr. Math Home || Search Dr. Math

```
Date: 05/22/2000 at 06:24:26
From: Cherriblossom

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

Thank you,
Cherriblossom
```

```
Date: 05/22/2000 at 07:06:35
From: Doctor Floor

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)

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

If you need more help, just write back.

Best regards,
- Doctor Floor, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Geometry
High School Triangles and Other Polygons

Search the Dr. Math Library:

 Find items containing (put spaces between keywords):   Click only once for faster results: [ Choose "whole words" when searching for a word like age.] all keywords, in any order at least one, that exact phrase parts of words whole words

Submit your own question to Dr. Math
Math Forum Home || Math Library || Quick Reference || Math Forum Search