Question

In a right angle triangle $\Delta ABC$, a perpendicular BD is drawn on to the largest side from the vertex B. Which of the following does NOT give the ratios of the areas of $\Delta ABD$ and $\Delta BCD$ ?

• A. ${\left(\frac{AB}{BC}\right)}^2$
• B. ${\left(\frac{AB}{AC}\right)}^2$
• C. ${\left(\frac{AD}{BD}\right)}^2$
• D. ${\left(\frac{BD}{DC}\right)}^2$

Answer 1

The correct option is B ${\left(\frac{AB}{AC}\right)}^2$

In $△ABC$,
$\angle A={90}^{\circ }-\angle C$

In $△BDC$,
$\angle DBC={90}^{\circ }-\angle C$
$\Rightarrow$ $\angle BAD=\angle DBC$

Consider $\Delta ABD$ and $\Delta BCD$,
$\angle BAD=\angle DBC$ (Proved above)
$\angle BDA=\angle BDC$ - - - - - -[Both are 90${}^{\circ }$ angles]
By AA similarity criterion, $\Delta ADB\sim \Delta BDC$
$\Rightarrow$ $\frac{Ar\left(\Delta ABD\right)}{Ar\left(\Delta BDC\right)}$
=${\left(\frac{AB}{BC}\right)}^2$
=${\left(\frac{AD}{BD}\right)}^2$
=${\left(\frac{BD}{DC}\right)}^2$
$\therefore$ $\frac{Ar\left(\Delta ABD\right)}{Ar\left(\Delta BDC\right)}$ $\ne$ ${\left(\frac{AB}{AC}\right)}^2$