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Question

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
465459.PNG

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Solution

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(BDBC)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=(BDBC)2(BC)2
Hence, Solving the above equation we get,
(AD)2=BD.DC

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