he proof is straightforward by using several triangle similarities of parallel lines crossing an angle. Let'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.
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.