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[几何] 等腰直角三角形的一个性质

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Zach posted 2024-6-14 09:05 |Read mode
若三角形$ABC$的角$C$为直角,且中线$AM$的斜率为$-2$,中线$BN$的斜率为$\frac{-1}{2}$,
证明;$AC=BC$,且$AC$是铅直线,$BC$是水平线

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Czhang271828 posted 2024-6-14 18:29
设中线交点是原点. 依照三等分关系, 设坐标
$$
A(2t,-4t),\quad M(-t,2t),\quad B(4s, -2s),\quad N(-2s,s) .
$$
由于 $M$ 是 $B$ 与 $C$ 的中点, 从而
$$
C(-2t-4s,4t+2s).
$$
将 $CA\perp CB$ 转化作斜率条件, 得
$$
0=(-2t-4s-2t,4t+2s+4t)\cdot (-2t-4s-4s,4t+2s+2s)=40(s+t)^2.
$$
因此 $s+t=0$. 从而 $A$ 与 $C$ 纵坐标相同, $B$ 与 $C$ 横坐标相同, $AB$ 斜率 $-1$.

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