# In the figure, ABD is triangle right angled at A and AC $$\perp$$ BD. Show that (i) $${AB}^2$$ = BC.BD (ii) $${AC}^2$$ = BC.DC (iii) $${AD}^2$$ = BD.CD

## Solution :

(i) Since, AC $$\perp$$ BD, therefore,

$$\triangle$$ ABC ~ $$\triangle$$ DBA  and  each triangle is similar to whole triangle ABD.

$$\implies$$  $$AB\over BD$$ = $$BC\over AB$$

So, $${AB}^2$$ = BC.BD

(ii) Since, $$\triangle$$ ABC ~ $$\triangle$$ DAC, therefore,

$$\implies$$  $$AC\over BC$$ = $$DC\over AC$$

So, $${AC}^2$$ = BC.DC

(iii) Since, $$\triangle$$ ACD ~ $$\triangle$$ BAD, therefore,

$$\implies$$  $$AD\over CD$$ = $$BD\over AD$$

So, $${AD}^2$$ = BD.CD