## Similar Triangles

In triangle $ABC, AB=20, BC=7$, and $CA=15$. Side $\overline{BC}$ is extended to point $D$ so that triangle $DAB$ is similar to triangle $DCA$. What is $DC$?

Source: NCTM Mathematics Teacher, November 2006

SOLUTION

Triangle $DAB$ is similar to triangle $DCA$
$\dfrac{DB}{DA}=\dfrac{DA}{DC}=\dfrac{AB}{CA}$

$\dfrac{DB}{DA}=\dfrac{DA}{DC}=\dfrac{4}{3}$

Solving $\dfrac{DA}{DC}=\dfrac{4}{3}$
$\dfrac{y}{x}=\dfrac{4}{3}$
Cross multiply
$3y=4x\qquad\qquad (1)$
Solving $\dfrac{DB}{DA}=\dfrac{4}{3}$
$\dfrac{7+x}{y}=\dfrac{4}{3}$
Multiply the left hand side by $3$
$\dfrac{3(7+x)}{3y}=\dfrac{4}{3}$
Substitute the value of $3y$ from Eq. $(1)$
$\dfrac{3(7+x)}{4x}=\dfrac{4}{3}$
Cross multiply
$9(7+x)=16x$
$63+9x=16x$
$63=7x$
$x=9$

Answer: $9$

Alternative solution
Triangle $DAB$ is similar to triangle $DCA$
$\dfrac{DC}{DA}=\dfrac{AD}{BD}=\dfrac{CA}{AB}$

$\dfrac{DC}{DA}=\dfrac{AD}{BD}=\dfrac{3}{4}$
$DC=(3/4)DA\qquad\qquad (2)$
$BD=(4/3)AD$
$BD-DC=(7/12)DA$
$BC=(7/12)DA$
$7=(7/12)DA$
$DA=12$
Substitute the value of $DA$ into Eq. $(2)$
$DC=(3/4)12=9$.