Question

∆ABD is a right triangle right-angled at A and AC ⊥ BD. Show that

(i) AB² = BC . BD
(ii) AC² = BC . DC
(iii) AD² = BD . CD
(iv) $\frac{{\mathrm{AB}}^2}{{\mathrm{AC}}^2}=\frac{\mathrm{BD}}{\mathrm{DC}}$

Answer 1

(i)

In and ,

$\angle ACB=\angle A=90°$

(Common angle)

So, by AA criterion

.....(1)

(ii) In and ,

(Common angle)

So, by AA criterion

.....(2)

(iii) We have shown that is similar to and is similar to therefore, by the property of transitivity is similar to .

.....(3)

(iv) Now to obtained AB²/AC² = BD/DC, we will divide equation (1) by equation (2) as shown below,

Canceling BC we get,

Therefore,