Trapezium

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Dec 18th 2010, 06:47 PM
rn5a

Trapezium

ABCD is a trapezium. How do I show that

1. a line parallel to the parallel sides of the trapezium cuts the non-paralle sides in the same ratio.

2. diagonals AC & BD in the trapezium cut each other in the same ratio.

Thanks,

Ron
Dec 19th 2010, 12:06 PM
earboth

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Quote:

Originally Posted by rn5a

ABCD is a trapezium. How do I show that

1. a line parallel to the parallel sides of the trapezium cuts the non-paralle sides in the same ratio.

2. diagonals AC & BD in the trapezium cut each other in the same ratio.

Thanks,

Ron

1. A trapezium is the bottom part of a triangle. Use the proportions of the lengths of the sides of the indicated triangles and the lengths of the thick black parallels.

2. Use the point of intersection as the vertex of triangles whose bases are parallel. Use proportions with the thick black parallels and the sides of the triangles indicated in similar colours.
Dec 19th 2010, 12:42 PM
Archie Meade

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Quote:

Originally Posted by rn5a

ABCD is a trapezium. How do I show that

1. a line parallel to the parallel sides of the trapezium cuts the non-paralle sides in the same ratio.

2. diagonals AC & BD in the trapezium cut each other in the same ratio.

Thanks,

Ron

Hi Ron,

if any line parallel to the sides [AB] and [DC] is drawn,

we can move side [BC] to the other end and focus on the similar triangles DAE and GAF.

DAE is a magnified version of DAF

$|AD|=m|AG|$

$|AE|=m|AF|$

$\displaystyle\ m=\frac{|AD|}{|AG|}=\frac{|AE|}{|AF|}$

For the diagonals, since [DC] and [AB] are parallel, triangles OCD and OAB are similar also.

$m|OC|=|OA|$

$m|OD|=|OB|$

$\displaystyle\ m=\frac{|OA|}{|OC|}=\frac{|OB|}{|OD|}$

Sorry about this Earboth....
my artwork was in progress before you posted (Cool)