Question

In the right triangle ΔABC, line segment CD is the median to the hypotenuse AB. then...
1)AB=DC; BD=DC
2)AD=DC; AC=DC
3)AD=DC; BD=DC
4)AB=DC; AC=DC​

Answer 1

Answer:

In △BCA and △BAD,

∠BCA=∠BAD        ....Each 90o

∠B is common between the two triangles.

So, △BCA∼△BAD        ...AA test of similarity     ....(I)

Hence, ABBC=ADAC=BDAB          ...C.S.S.T

And, ∠BAC=∠BDA           ....C.A.S.T        ....(II)

So, ABBC=BDAB

∴AB2=BC×BD

Hence proved.

(ii) In △BCA and △DCA,

∠BCA=∠DCA        ....Each 90o

∠BAC=∠CDA        ...From (II)

So, △BCA∼△ACD        ...AA test of similarity         ....(III)

Hence, ACBC=CDAC=ADAB          ...C.S.S.T

So, ACBC=CDAC

∴AC2=BC×DC

Hence proved.

(iii) From (I) and (III), we get

△BAD∼△ACD

Hence, ACAB