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Friday, 4 December 2015

Question: Suppose that the sides AB, BC, CD and DA of a quadrilateral ABCD are cut by a line at the points A' B' C' D' respectively, show that AA'/A'B \times BB'/B'C \times CC'/C'D \times DD'/D'A = +1

Question (harder geometry):
Suppose that the sides AB, BC, CD and DA of a quadrilateral ABCD are cut by a line at the points A' B' C' D' respectively, show that 
AA'/A'B \times BB'/B'C \times CC'/C'D \times DD'/D'A = +1

SOLUTION:  ================


Referring to my diagram, see attached.

I simply used the sine rule 4 times on triangles A'BB', CB'C', C'DD', AA'D ;

let the angles be \alpha, \beta, \gamma, \delta.

A'B/\sin(\pi-\alpha-\beta) = BB'/\sin \alpha (1)
C'D/\sin \delta = DD'/sin (\pi-\gamma-\delta) (2)
B'C/\sin(\pi-\gamma-\delta) = CC'/\sin(\pi-\alpha-\beta) (3)
AA/\sin(\pi-\delta) = AD'/\sin (\pi-\alpha) (4)

so
BB'/A'B = \sin \alpha/ \sin(\pi-\alpha-\beta)
DD'/C'D = \sin (\pi-\gamma-\delta)/\sin \delta
CC'/B'C = \sin(\pi-\alpha-\beta)/\sin(\pi-\gamma-\delta)
AA'/AD' = \sin(\pi-\delta)/\sin(\pi-\alpha) = \sin \delta/sin \alpha
[since \sin(A-B)=\sin A \cos B - \cos B \sin A, and \cos \pi = -1]

Then we have

AA'/A'B \times BB'/B'C \times CC'/C'D \times DD'/D'A
= BB'/A'B\times CC'/B'C\times DD'/C'D\times AA'/D'A
={\sin\alpha \over \sin(\pi-\alpha-\beta)}\times {\sin(\pi-\alpha-\beta)\over \sin(\pi-\gamma-\delta)}\times {\sin (\pi-g-\delta)\over \sin \delta}\times {\sin \delta\over \sin \alpha}

= 1 [since everything cancels to 1.]
as required.//

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