Question 3:

In the following figure, ABD is a triangle right angled at A and AC ⊥ BD. Show that

(i) AB² = BC × BD
(ii) AC² = BC × DC
(iii) AD² = BD × CD

Answer:

(i) In ∆ADB and ∆CAB,

∠DCA = ∠ DAB (Each 90º)

∠CDA = ∠ ADB (Common angle)

∴∆DCA ~ ∆DAB

⇒ DC/DA = DA/DB

⇒ AD² = BD × CD

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