Fc Năng Lượng Hạt Nhân
11/12/2017 at 21:06-
KEITA FC 8C 11/12/2017 at 21:21
Consider trapezoidal trapezoid trapezoid ABCD (AB parallelogram CD) \(BH=\dfrac{AB+CD}{2}\) \(\left(1\right)\)
Pass straight line parallel with AC, cut CD at E .
We have : BE = AC , AC = BD Candles BE = BD .
Triangle BDE weighs in at B road BH should \(DH=HE=\dfrac{DE}{2}\) \(\left(2\right)\)
We have : AB = CE Candles \(AB+CD=CE+CD=DE\) \(\left(3\right)\)
From ( 1 ) , ( 2 ) , ( 3 ) \(\Rightarrow BH=DH=HE\)
The triangle BHD, BHE square weight at B should angle DBE = 90 degrees
We have : DB perpendicular to BE, AC in parallel with BE so DB perpendicular to AC .
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