Q22 of 95 Page 6

In the given figure, ABD is a right angled triangle being right angled at A and AD ⊥ BC. Show that:


(i) AB2= BC.BD


(ii) AC2 = BC. DC


(iii) AB. AC. = BC. AD

(i) In ΔDAB and ΔACB


∠ADB = ∠CAB [each 90°]


∠DAB = ∠CAB [common angle]


∴ DAB ~ACB [by AA similarity]


Since the triangles are similar, hence corresponding sides are in proportional.



⇒ AB2= BC×BD


(ii) In ACB and DAC


∠CAB = ∠ADC [each 90°]


∠CAB = ∠CAD [common angle]


∴ ACB ~DAC [by AA similarity]


Since the triangles are similar, hence corresponding sides are in proportional.



⇒ AC2 = BC. DC


(iii) In part (i) we proved that DAB ~ACB



⇒ AB × AC = BC × AD


Hence Proved


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