perplexus.info :: Geometry : Similar Triangles

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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