Question

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

Answer 1

Consider $△$ADB and $△$ABC

∠BAD = ∠BAC [common angle]

∠BDA = ∠ABC [ ${90}^{\circ }$]

Therefore by AA similarity criterion,
$△$ADB and $△$ABC are similar. (1 mark)

So, $\frac{AB}{AC}$ = $\frac{AD}{AB}$

⇒ ${AB}^2$ = AC. AD -------------(I)

Similarly, $△$BDC and $△$ABC are similar.

So, $\frac{BC}{AC}$ = $\frac{DC}{BC}$

⇒ ${BC}^2$ = AC. DC -------------(II) (1 mark)

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. (1 mark)