Question

$△$ABC is a right angled triangle, right angled at B. BD is perpendicular as shown. If AD: DC = 1:2, then $\frac{{AB}^2}{{BC}^2}$ is:

• A. 1: 2
• B. 1: 4
• C. 4: 1
• D. 2: 1

Answer 1

The correct option is A

1: 2

Consider $△$ADB and $△$ABC
∠BAD = ∠BAC [common angle]
∠BDA = ∠ABC [ ${90}^{\circ }$]
Therefore by AA similarity criterion, $△$ADB and $△$ABC are similar.
So, $\frac{AB}{AC}$ = $\frac{AD}{AB}$
$\Rightarrow$ ${AB}^2$ = AC . AD -------------(I)

Similarly, $△$BDC and $△$ABC are similar.
So, $\frac{BC}{AC}$ = $\frac{DC}{BC}$
$\Rightarrow$ ${BC}^2$ = AC . DC -------------(II)
Dividing (I) and (II) and cancelling out AC, we get, $({\frac{AB}{BC})}^2$ = $\frac{AD}{DC}$ -------------(III)
So, the ratio is the same, i.e., 1: 2.