All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > Shapes > Geometry
Similar Triangles (Posted on 2013-11-15) Difficulty: 3 of 5

  
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(5): Argand - grand! | Comment 9 of 10 |
(In reply to re(4): Argand - grand! by Harry)



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

More detail on "Construct ray BP" such that /ABP" = /AFC".

1) If any of the rays BA, FA, or FC are too short you should extend them.
2) Construct circle(F,r) intersecting FA at K and FC at L.
3) Construct circle(B,r) intersecting BA at I.
4) Construct circle(I,|KL|) intersecting circle(B,r) at two points and pick one as P" such that /ABP" = /AFC as directed angles.

Note that when constructing the circles that only enough of the circle is needed to define any intersection points.

Hope this helps.

  Posted by Bractals on 2013-12-04 05:14:19
Please log in:
Login:
Password:
Remember me:
Sign up! | Forgot password


Search:
Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (6)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2017 by Animus Pactum Consulting. All rights reserved. Privacy Information