Ratios in a right triangle

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Ratios in a right triangle

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#1 h Jun 9, 2023, 10:58 PM

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Let $ABC$ be a right triangle in $A$ with $AB<AC$ . Let $M$ be the midpoint of $AB$ and $D$ a point on $AC$ such that $DC=DB$ . Let $X=(BDC)\cap MD$ .
Compute in terms of $AB,BC$ and $AC$ the ratio $\frac{BX}{DX}$ .

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#2 h Apr 6, 2025, 1:15 PM

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In the $xy-$ plane let $A(0,0); B(0,c); C(b,0); M(0,c/2)$ (with $b>c$ ) be defined for right $\Delta ABC$ . Now, let $D(t,0)$ lie on the same line as $AC$ and to the left of $\Delta ABC$ . If $|DB|=|DC|$ , then:

$\sqrt{t^2+c^2}=b+t \Rightarrow t^2+c^2 = t^2+2bt+b^2 \Rightarrow t = -\frac{b^2-c^2}{2b}$ (i)

Extending the line $MD$ to $BC$ results in the point $X = MD \cap \Delta BDC$ , or:

$MD: y - c/2 = \frac{c/2}{\frac{b^2-c^2}{2b}}x \Rightarrow y = \frac{bc}{b^2-c^2}x + \frac{c}{2}$ (ii);

$BC: y = -\frac{c}{b}x + c$ (iii);

or $X(x,y) = \left(\frac{(b^2-c^2)b}{2b^2-c^2}, \frac{b^2c}{2b^2-c^2}\right)$ . This leads to the computations of $|BX|, |DX|$ :

$|BX| = \sqrt{\left(0-\frac{(b^2-c^2)b}{2b^2-c^2}\right)^2 + \left(c-\frac{b^2c}{2b^2-c^2}\right)^2} = \left(\frac{b^2-c^2}{2b^2-c^2}\right)\sqrt{b^2+c^2}$ (iv);

$|DX| = \sqrt{ \left( -\frac{b^2-c^2}{2b} - \frac{(b^2-c^2)b}{2b^2-c^2} \right)^{2} + \left(0 - \frac{b^{2}c}{2b^2-c^2} \right)^{2} } = \frac{\sqrt{4b^{6}c^{2} +(4b^2-c^2)^{2}(b^2-c^2)^{2} } } {2b(2b^2-c^2)}$ (v);

of which:

$\frac{|BX|}{|DX|} = \frac{2b(b^2-c^2)\sqrt{b^2+c^2}}{\sqrt{4b^{6}c^{2}+(4b^2-c^2)^{2}(b^2-c^2)^{2}}} = \textcolor{red}{\frac{2|AC||BC|(|AC|^2-|AB|^2)}{\sqrt{4|AC|^{6}|AB|^{2} +(4|AC|^2-|AB|^2)^{2}(|AC|^2-|AB|^2)^{2}}}}$ .

This post has been edited 7 times. Last edited by Mathzeus1024, Apr 6, 2025, 1:45 PM

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