In Fig., 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

In △DCA, By Pythagoras theorem,
(DA)²=(CA)²+(CD)².. (1)
In △ABD, By Pythagoras theorem,
(BD)²=(AB)²+(AD)² (2)
In △ABC,
(i)By Pythagoras theorem,
(AB)²=(AC)²+(BC)².. (3)
∴(AB)²=(AD)²−(CD)²+(BC)² (From 1 and 3)
∴(AB)²=(BD)²−(AB)²−(CD)²+(BC)²
∴2(AB)²=(BD)²−(BD−BC)²+(BC)²
∴(AB)²=BC.BD
(ii)Adding (1) and (3),
∴(AB)²+(AD)²=2(AC)²+(CD)²+(BC)²
∴(BD)²−(AD)²+(AD)²=2(AC)²+(CD)²+(BC)² (From 2)
∴(BC+CD)²=2(AC)²+(CD)²+(BC)²
Hence, Solving the above equation we get,
(AC)²=BC.DC
(iii)Subtracting (1) and (2) we get,
2(AD)²−(BD)²=(CA)²+(CD)²−(AB)²
∴2(AD)²−(BD)²=(CA)²+(CD)²−(AC)²−(BC)²
∴2(AD)²−(BD)²=(BD−BC)²−(BC)²
Hence, Solving the above equation we get,
(AD)²=BD.DC