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

[Albert1]

$\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 $\overline{BD}: \overline{CD}$

 

[Greg]

Spoiler

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

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

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

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

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

\(\displaystyle 7x^2-7y^2+30y-175=0\)

\(\displaystyle 7(7-y)^2-7y^2+30y-175=0\Rightarrow y=\frac{42}{17},x=7-y=\frac{77}{17},\frac xy=\frac{11}{6},\overline{BD}:\overline{CD}=11:6\)

 

[Albert1]

Spoiler

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

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

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

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

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

\(\displaystyle 7x^2-7y^2+30y-175=0\)

\(\displaystyle 7(7-y)^2-7y^2+30y-175=0\Rightarrow y=\frac{42}{17},x=7-y=\frac{77}{17},\frac xy=\frac{11}{6},\overline{BD}:\overline{CD}=11:6\)

very good , your answer is right!