Seg AD is the median of ABC and AM - BC- Prove thati AC 2- AD 2- BC - DM - BC 22 ii AB 2- AD 2- BC - DM - BC 22Note- ii part should be AB 2- AD 2- BC - DM - BC 22

Note: (ii) part should be AB2=AD2 BC×DM+BC22 (i) nbsp;In nbsp;△ADC,∠ADC gt;90° nbsp; nbsp; nbsp; nbsp; nbsp; nbsp; nbsp; nbsp; nbsp; ...

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Pythagoras Theorem

Seg AD is the...

Question

$Seg AD is the median of △ ABC and AM ⊥ BC . Prove that (i) {AC}^{2} = {AD}^{2} + BC \times DM + {(\frac{BC}{2})}^{2} (ii) {AB}^{2} = {AD}^{2} + BC \times DM + {(\frac{BC}{2})}^{2}$

Note: (ii) part should be ${AB}^{2} = {AD}^{2} - BC \times DM + {(\frac{BC}{2})}^{2}$

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△ A B C

⊥

A C^{2} = A D^{2} + B C \times D M + \frac{B C^{2}}{4}

A B^{2} = A D^{2} - B C \times D M + \frac{B C^{2}}{4}

A D

A B C

A M ⊥ B C .

(i) A C^{2} = A D^{2} + B C \times D M + {(\frac{B C}{2})}^{2}

A D

A B C

A M ⊥ B C .

(i i) A B^{2} = A D^{2} - B C \times D M + {(\frac{B C}{2})}^{2}

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(i i i) A C^{2} + A B^{2} = 2 A D^{2} + \frac{1}{2} B C^{2}

In the given figure, AD is a median of a triangle ABC and AM ⊥ BC. Prove that:

(i)

(ii)

(iii)

A D

A B C

A M ⊥ B C

A D^{2} - B C . D M + {[\frac{B C}{2}]}^{2} =

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