Find the Ratio of $\overline{BD}: \overline{CD}$ in $\triangle ABC$

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SUMMARY

The discussion focuses on finding the ratio of segments $\overline{BD}$ to $\overline{CD}$ in triangle $\triangle ABC$ where $\overline{AB}=5$, $\overline{AC}=6$, and $\overline{BC}=7$. The point $M$ is defined as the midpoint of $\overline{AC}$, and point $D$ lies on $\overline{BC}$ such that $\overline{AD}$ is perpendicular to $\overline{BM}$. The conclusion reached is that the ratio $\overline{BD}:\overline{CD}$ is 5:2.

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$\triangle ABC, \overline{AB}=5,\overline{AC}=6,\overline{BC}=7,$$M$ is the midpoint of $\overline{AC}$
$D$ is one point on $\overline{BC}$,if $\overline{AD}\perp \overline{BM}$,
please find the ratio of $\overline{BD}: \overline{CD}$
 

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$$5^2=7^2+6^2-2(7)(6)\cos C\Rightarrow\cos C=\frac57$$

In an orthodiagonal quadrilateral, the sum of the squares of opposite sides are equal, hence

$$x^2+3^2=\overline{MD}^2+5^2\Rightarrow x^2-\overline{MD}^2-16=0\quad[1]$$

$$\overline{MD}^2=9+y^2-2(3)(y)\cos C\Rightarrow\overline{MD}^2=9+y^2-\frac{30}{7}y\quad[2]$$

Substituting $$[2]$$ into $$[1]$$ gives $$x^2- y^2+\frac{30}{7}y-25=0$$

$$7x^2-7y^2+30y-175=0$$

$$7(7-y)^2-7y^2+30y-175=0\Rightarrow y=\frac{42}{17},x=7-y=\frac{77}{17},\frac xy=\frac{11}{6},\overline{BD}:\overline{CD}=11:6$$
 
greg1313 said:
$$5^2=7^2+6^2-2(7)(6)\cos C\Rightarrow\cos C=\frac57$$

In an orthodiagonal quadrilateral, the sum of the squares of opposite sides are equal, hence

$$x^2+3^2=\overline{MD}^2+5^2\Rightarrow x^2-\overline{MD}^2-16=0\quad[1]$$

$$\overline{MD}^2=9+y^2-2(3)(y)\cos C\Rightarrow\overline{MD}^2=9+y^2-\frac{30}{7}y\quad[2]$$

Substituting $$[2]$$ into $$[1]$$ gives $$x^2- y^2+\frac{30}{7}y-25=0$$

$$7x^2-7y^2+30y-175=0$$

$$7(7-y)^2-7y^2+30y-175=0\Rightarrow y=\frac{42}{17},x=7-y=\frac{77}{17},\frac xy=\frac{11}{6},\overline{BD}:\overline{CD}=11:6$$
very good , your answer is right!
 

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