Cross's Theorem.

mfgoode · 2005-07-09, 16:38

This theorem was discovered by a schoolboy-David Cross
Squares have been drawn on the sides of an arbitrary triangle ABC and the free corners joined to make three more triangles.
The areas of these new triangles are all (each ) equal to the area of the original triangle. WHY?

Mally

tom11784 · 2005-07-10, 01:43

Let's say we have a triangle with sides a,b,c and opposite angles A,B,C.
We know that the area of the original triangle equals all of
.5ab sinC, .5bc sinA, .5ca sinB

Let's look at the triangle created from the squares off of sides a & b
The triangle has two sides with length a and b, and the angle between them equals 180-C deg
Thus the area of the new triangle equals .5ab sin(180-C) = .5ab sinC for C<180
The same applies for the triangles between the other pairs of squares.

mfgoode · 2005-07-10, 16:12

Very good Tom 11784. The deduction of the angle (180 - C) is a good piece of deduction. You are very strong in Trig. I notice
Try this alternate method. The proof rests on the Theorem that triangles on equal bases and same height are equal in area. This condition can be met by simply rotating a square by 90 degrees. If you dont get it I'll show you.
With the absence of a diagram its difficult to explain wthout suitable notation Cheers
Mally

2005-07-09, 16:38	#1
mfgoode Bronze Medalist Jan 2004 Mumbai,India 2²·3³·19 Posts	Cross's Theorem. This theorem was discovered by a schoolboy-David Cross Squares have been drawn on the sides of an arbitrary triangle ABC and the free corners joined to make three more triangles. The areas of these new triangles are all (each ) equal to the area of the original triangle. WHY? Mally

2005-07-10, 16:12	#3
mfgoode Bronze Medalist Jan 2004 Mumbai,India 2²×3³×19 Posts	Cross's Theorem Very good Tom 11784. The deduction of the angle (180 - C) is a good piece of deduction. You are very strong in Trig. I notice Try this alternate method. The proof rests on the Theorem that triangles on equal bases and same height are equal in area. This condition can be met by simply rotating a square by 90 degrees. If you dont get it I'll show you. With the absence of a diagram its difficult to explain wthout suitable notation Cheers Mally

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2005-07-10, 01:43	#2
tom11784 Aug 2003 Upstate NY, USA 2·163 Posts	Let's say we have a triangle with sides a,b,c and opposite angles A,B,C. We know that the area of the original triangle equals all of .5ab sinC, .5bc sinA, .5ca sinB Let's look at the triangle created from the squares off of sides a & b The triangle has two sides with length a and b, and the angle between them equals 180-C deg Thus the area of the new triangle equals .5ab sin(180-C) = .5ab sinC for C<180 The same applies for the triangles between the other pairs of squares.