In fig. 6.53,ABD is a triangle right angled at A and AC | BD Show that
• AB² = BC.BD • AC²= BC .BD • AD²= BD.CD
Answers
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Step-by-step explanation:
i) In △BCA and △BAD,
∠BCA=∠BAD ....Each 90
o
∠B is common between the two triangles.
So, △BCA∼△BAD ...AA test of similarity ....(I)
Hence,
AB
BC
=
AD
AC
=
BD
AB
...C.S.S.T
And, ∠BAC=∠BDA ....C.A.S.T ....(II)
So,
AB
BC
=
BD
AB
∴AB
2
=BC×BD
Hence proved.
(ii) In △BCA and △DCA,
∠BCA=∠DCA ....Each 90
o
∠BAC=∠CDA ...From (II)
So, △BCA∼△ACD ...AA test of similarity ....(III)
Hence,
AC
BC
=
CD
AC
=
AD
AB
...C.S.S.T
So,
AC
BC
=
CD
AC
∴AC
2
=BC×DC
Hence proved.
(iii) From (I) and (III), we get
△BAD∼△ACD
Hence,
AC
AB
=
CD
AD
=
AD
BD
So, AD
2
=BD×CD
Hence proved.
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