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Xét hình thang cân ABCD có đáy nhỏ AB = m, đáy lớn CD = n, đường cao AH = BK = h, DH = KC = a ---> CD = n = m + 2a
Theo định lý Pythagore, ta có :
AC^2 = AH^2 + HC^2 (1)
AD^2 = AH^2 + DH^2 (2)
---> AC^2 - AD^2 = HC^2 - DH^2 = (m+a)^2 - a^2 = m^2 + 2m.a = m(m+2a) = m.n = AB.CD (đpcm)
đơn giản như đan rổ