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 Similar Triangles (Posted on 2013-11-15)

Let ABCD be a simple quadrilateral.

Construct with straightedge and compass a point P
such that ΔPAB ~ ΔPCD and both triangles have the
same orientation.

 See The Solution Submitted by Bractals Rating: 4.0000 (1 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
 re: Argand - grand! | Comment 5 of 10 |
(In reply to Argand - grand! by Harry)

No. the following constructon is standard.

Construct ray AP' such that /BAP' = /FAC.
Construct ray BP" such that /ABP" = /AFC.
P is the intersection of AP' and BP".

Bye the way, how would you solve the problem if the triangles should have opposite orientation.

 Posted by Bractals on 2013-11-28 22:36:03
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