Let a, b, c, d ratios of circles having points A,B,C,D, 2L the length of the segment passing through centers and E and F the centers of circles a and d. Ratios are related and it's easy to get: b=L-c; a=(L+c)/2; d=L-c/2
Apply Pythagora's theorem in triangles QKE and PHF to prove AB=CD.
If we complete lower semi-circles and trace CD tangent to the upper ones, we can also prove that AC=BD.
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Let a, b, c, d ratios of circles having points A,B,C,D, 2L the length of the segment passing through centers and E and F the centers of circles a and d. Ratios are related and it's easy to get:
b=L-c; a=(L+c)/2; d=L-c/2
Apply Pythagora's theorem in triangles QKE and PHF to prove AB=CD.
If we complete lower semi-circles and trace CD tangent to the upper ones, we can also prove that AC=BD.
César Lozada
Venezuela
Sorry. I forgot to mention QK perpendicular to AE and PH perpendicular to FD
ReplyDeleteCésar Lozada
Venezuela
Prove: AB=CD=sqrt(2bc)
ReplyDeleteCésar Lozada
Venezuela
You can also use a=c+b/2 and d=b+c/2
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