Geometric Art: Typography, iPad

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## Wednesday, February 21, 2018

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## Wednesday, February 21, 2018

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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.