Question

ABD is a triangle right angled at A and AC 1 BD Show that AB2=bc.bd​

Answer 1

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.