Bốn điểm phân biệt $A, B, C, D$ trên trục $x'Ox$ được gọi là một hàng điểm điều hòa khi: $\frac{\overline{CA} }{\overline{CB} }=-\frac{\overline{DA} }{\overline{DB} }$, kí hiệu $(ABCD)=-1$. Chứng minh: $(ABCD)=-1\Leftrightarrow \frac{2}{\overline{AB}}=\frac{1}{\overline{AC} }+\frac{1}{\overline{AD} }.$
Gọi $a, b, c, d$ lần lượt là tọa độ của $A, B, C, D$
Ta có: $\frac{\overline{CA} }{\overline{CB} }=-\frac{\overline{DA} }{\overline{DB} }$
$\Leftrightarrow (c-a)(b-d)=-(b-c)(a-d)$
$\Leftrightarrow ac+bc+ad+bd=2ab+2cd\Leftrightarrow (a+b)(c+d)=2(ab+cd)$
Và $\frac{2}{\overline{AB}}=\frac{1}{\overline{AC} }+\frac{1}{\overline{AD} }\Leftrightarrow \frac{2}{b-a}=\frac{1}{c-a}+\frac{1}{d-a}$
$\Leftrightarrow 2(c-a)(d-a)=(b-a)(d-a+c-a)$
$\Leftrightarrow (a+b)(c+d)=2(ab+cd)$. Từ đó suy ra đpcm.
Đặc biệt, nếu chọn $A$ làm gốc mới thì $a=0$ nên có lời giải gọn hơn:
$(ABCD)=-1\Leftrightarrow\frac{\overline{CA} }{\overline{CB} }=-\frac{\overline{DA} }{\overline{DB} }\Leftrightarrow b(c+d)=2cd$
và $\frac{2}{\overline{AB}}=\frac{1}{\overline{AC} }+\frac{1}{\overline{AD} }\Leftrightarrow 2cd=b(c+d)$: đúng

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