In the given figure- AD is a median of a triangle ABC and AM - BC- Prove that-i AC 2- AD 2- BC - DM - BC -22ii AB 2- AD 2- BC - DM - BC -22iii AC 2- AB 2-2 AD 2-1-2 BC 2

(i) Applying Pythagoras theorem in ΔAMD, we obtain AM2 + MD2 = AD2 … (1) Applying Pythagoras theorem in ΔAMC, we obtain AM2 + MC2 = AC2 AM2 + (MD + DC)2 = ...

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Question

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

(i)

(ii)

(iii)

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Solution

(i) Applying Pythagoras theorem in ΔAMD, we obtain

AM² + MD² = AD² … (1)

Applying Pythagoras theorem in ΔAMC, we obtain

AM² + MC² = AC²

AM² + (MD + DC)² = AC²

(AM² + MD²) + DC² + 2MD.DC = AC²

AD² + DC² + 2MD.DC = AC² [Using equation (1)]

Using the result, , we obtain

(ii) Applying Pythagoras theorem in ΔABM, we obtain

AB² = AM² + MB²

= (AD² − DM²) + MB²

= (AD² − DM²) + (BD − MD)²

= AD² − DM² + BD² + MD² − 2BD × MD

= AD² + BD² − 2BD × MD

(iii)Applying Pythagoras theorem in ΔABM, we obtain

AM² + MB² = AB² … (1)

Applying Pythagoras theorem in ΔAMC, we obtain

AM² + MC² = AC² … (2)

Adding equations (1) and (2), we obtain

2AM² + MB² + MC² = AB² + AC²

2AM² + (BD − DM)² + (MD + DC)² = AB² + AC²

2AM²+BD² + DM² − 2BD.DM + MD² + DC² + 2MD.DC = AB²+ AC²

2AM² + 2MD² + BD² + DC² + 2MD (− BD + DC) = AB² + AC²

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Similar questions

Q. (1)
In Figure,

A D

is a median of a triangle

A B C

and

A M ⊥ B C .

Prove that :

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

(2)
In Figure,

A D

is a median of a triangle

A B C

and

A M ⊥ B C .

Prove that :

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

(3)
In Figure,

A D

is a median of a triangle

A B C

and

A M ⊥ B C .

Prove that :

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

Q. Seg AD is the median of

△ A B C

and AM

⊥

BC.

Prove that:

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

ii)

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

Q. If in

Δ A B C

A D

is median and

A M ⊥ B C

, then prove that

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

Q. If in

△ A B C, A D

is median and

A E ⊥ B C

, then prove that

A B^{2} + A C^{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}

Note: (ii) part should be

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

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

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

(i)

(ii)

(iii)