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Given Right triangle ABC (right angle at C ) and DE perpendicular to AB at E. Prove triangle ABC is similar to triangle DBE. Use isometry. Please, help
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|dw:1446863249304:dw|
AAA to prove similar triangles so it is right there :)
I have to use isometry, I cannot use congruence or similarity.
That is I have to find out a rotation or dilation or both to make the graph look similar
isometry as in the ratios of the 3 sides are the same right? i think that is how to prove similar triangles w isometry...
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oh nvd....hmmm.....
Like this, if I "cut" triangle BDE out and "stand" the triangle ABC up, I have: |dw:1446870093138:dw|
|dw:1446870164799:dw| will this work? dilate BD'E' from there...
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