ICSE Class 9 Answered
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 | 08 Jun, 2020, 19:35: PM

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 | 09 Jun, 2020, 09:45: AM
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