Chủ đề này có 2 trả lời

[I Love MC]

Tam giác  ABC có AB=AC, góc A nhọn, BH vuông góc AC. C/minh $\frac{AH}{HC}=2.\frac{AB^2}{BC^2} -1$

[Phung Quang Minh]

Tam giác  ABC có AB=AC, góc A nhọn, BH vuông góc AC. C/minh $\frac{AH}{HC}=2.\frac{AB^2}{BC^2} -1$

-Ta có: \[A{B^2} - A{H^2} = B{C^2} - H{C^2}( = B{H^2})\] \[ =  > A{B^2} - B{C^2} = A{H^2} - H{C^2} = (AH - HC)(AH + HC) = (AH - HC).AC\].

=> \[AH - HC = \frac{{A{B^2} - B{C^2}}}{{AB}};AH + HC = AC = \frac{{A{B^2}}}{{AB}} =  > 2AH = \frac{{2A{B^2} - B{C^2}}}{{AB}};2HC = \frac{{B{C^2}}}{{AB}}\]

=> \[\frac{{AH}}{{HC}} = \frac{{\frac{{2A{B^2} - B{C^2}}}{{AB}}}}{{\frac{{B{C^2}}}{{AB}}}} = \frac{{2A{B^2} - B{C^2}}}{{B{C^2}}} = \frac{{2A{B^2}}}{{B{C^2}}} - 1.\]

[Quoc Tuan Qbdh]

- Perfect =)) -

$\frac{AH}{HC}=2.\frac{AB^{2}}{BC^{2}}-1\Leftrightarrow \frac{AH+HC}{HC}=\frac{2AB^{2}}{BC^{2}}\Leftrightarrow \frac{1}{HC}=\frac{2AB}{BC^{2}}\rightarrow 2AB.HC=HC^{2}+HB^{2}$