State true or false- In a - ABC- AD is median and AL BCthen AC2 - AD2 - BC-DL - 2 BC-22

The correct option is A TrueIn right angled triangle ALC, AC^2=AL^2+LC^2 #160; #160; #160; ........ (1) In right angled triangle ALD, ...

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Pythagoras Theorem

State true or...

Question

State true or false.
$I n a Δ A B C, A D$ is median and $A L ⊥ B C$ then
$A C^{2} = A D^{2} + B C . D L + 2 {(\frac{B C}{2})}^{2}$

True

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False

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Solution

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(i) {AC}^{2} = {AD}^{2} + BC . DL + {(\frac{BC}{2})}^{2} (ii) {AB}^{2} = {AD}^{2} - BC . DL + {(\frac{BC}{2})}^{2} (iii) {AB}^{2} + {AC}^{2} = 2 {AD}^{2} - \frac{1}{4} {BC}^{2}

△ A B C

A D

B C

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

△ 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

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

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

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

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}

{AB}^{2} = {AD}^{2} - BC \times DM + {(\frac{BC}{2})}^{2}

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