We observe from Fig 22 that
AC²
AB + BC² = 1² + 1²
AC³= 2 √2
AC
AD² = AC²+CD² = (√2)² + 1² = 2+1=3
AD
Similarly, we get the other lengths
Answers
Answer:
(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
Answer:
(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.