# 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, 07:35: PM

Expert Answer

In right ABC, using Pythagoras theorem, we have:

AC^{2} = AB^{2} + BC^{2}

AB^{2} = AC^{2} - BC^{2} ...(i)

In right ABD, using Pythagoras theorem, we have:

AD^{2} = AB^{2} + BD^{2}

AB^{2} = AD^{2} - BD^{2} ... (ii)

1.

From (i) and (ii), we have

AD^{2} - BD^{2} = AC^{2} - BC^{2}

AD^{2} - BD^{2} = AC^{2} - (BD + DC)^{2}

AD^{2} - BD^{2} = AC^{2} - BD^{2} - DC^{2} - 2BD x DC

AD^{2} - BD^{2} = AC^{2} - BD^{2} - DC^{2} - 2DC^{2} ... (BD = DC)

AD^{2} = AC^{2} - 3DC^{2}

AC^{2} = AD^{2} + 3DC^{2}

2.

From (ii), we have

AD^{2} = AB^{2} + BD^{2}

BC^{2} = 4AD^{2} - 4AB^{2}

BC^{2} = 4(AD^{2} - AB^{2}) ... (ii)

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

AC^{2} = 4AD^{2} - 3AB^{2}

Hence, proved.

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

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