Geometric Art: Typography, iPad
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Wednesday, February 21, 2018
Typography of Geometry Problem 865: Parallelogram, Diagonal, Midpoint, Side, Triangle, Parallel, Similarity, iPad Apps
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he proof is straightforward by using several triangle similarities of parallel lines crossing an angle.
ReplyDeleteLet's call h the height from H to AD.
Let's call y the distance between parallel lines AD and OM.
y is also the distance between lines OM and BC.
Let's call g the height from G to AD.
Let's call s=AF=FD=BE=EC.
K is the intersection between OM and GF.
L is the interesection between OM and GD.
We have that:
1. OK=OL-KL=s-KL
We have the following relations from similar triangles:
2. KL=s×(g-y)/g
3. DG/g=OE/y
4. OE/DG=OH/DH=(h-y)/h
5. OM=s×(h-y)/h
From (1-4) we have that:
6. OK=OM=s×(h-y)/h.
(6) implies that MOE is congruent to KOF, which in turn means that ME is parallel to FG. Q.E.D.