triangle cde excists with in triangle cba and is similar to triangle cba with Angle cde congruent to angle b. if cd= 10, ca = 16, and eb = 12, find ce?
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triangle cde excists with in triangle cba and is similar to triangle cba with Angle cde congruent to angle b. if cd= 10, ca = 16, and eb = 12, find ce?
If I interpret your question correctly (which is not easy when you use lower case letters), ?CDE ~ ?CBA. So CD:CB = DE:BA = CE:CA. (These are proportions from the definition: Corresponding sides of similar triangles are proportional.)
Substituting the numbers in: 10/(x + 12) = x/16.
I think you may have written the triangles' names in improper order; always arrange corresponding vertices in the same order. I would have written ?CDE ~ ?CAB myself, so that CD:CA = CE:CB and 10/16 = x/(x + 12)
Then using the Means-Extremes Property or "cross multiplying", 10(x + 12) = 16x
10x + 120 = 16x
120 = 6x
x = 20
CE = x = 20
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