# 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

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 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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