in tringle ABC, angle B is 90 °and D is the mid point of BC . prove that 

1. AC² = AD² + 3CD²

2.BC²= 4(AD² -AB²)


 

 

Asked by kavyaraval1075.9sdatl | 8th Jun, 2020, 07:35: PM

Expert Answer:

In right ABC, using Pythagoras theorem, we have:

AC2 = AB2 + BC2 

AB2 = AC2 - BC2 ...(i)

In right ABD, using Pythagoras theorem, we have:

AD2 = AB2 + BD2 

AB2 = AD2 - BD2 ... (ii)

1.

From (i) and (ii), we have

AD2 - BD2 = AC2 - BC2

AD2 - BD2 = AC2 - (BD + DC)2

AD2 - BD2 = AC2 - BD2 - DC2 - 2BD x DC

AD2 - BD2 = AC2 - BD2 - DC2 - 2DC2  ... (BD = DC)

AD2 = AC2 - 3DC2

AC2 = AD2 + 3DC2 

2.

From (ii), we have

AD2 = AB2 + BD2 

 

BC2 = 4AD2 - 4AB2 

BC2 = 4(AD2 - AB2) ... (ii)

Substituting (ii) in (i), we get,

AC2 = 4AD2 - 3AB2

Hence, proved.

 

Answered by Renu Varma | 9th Jun, 2020, 09:45: AM