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)

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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.