Question

Plz help me .....





.in solving this question

Answer 1

Step-by-step explanation:

We have,

∠BAC = 90°, seg AD, BE and CF are medians.

(i) In ΔBAC:

⇒ AB² + AC² = 2AD² + 2BD²

⇒ AB² + AC² - 2BD² = 2AD²

⇒ AB² + AC² - (1/2) BC² = 2AD²

⇒ 2AB² + 2AC² - BC² = 4AD²

Likewise,

⇒ 2AB² + 2BC² - AC² = 4BE²    ------ (ii)

 

⇒ 2AC² + 2BC² - AB² = 4CF²     ------ (iii)

On solving (i), (ii) & (iii), we get

⇒ 2AB² + 2AC² - BC² + 2AB² + 2BC² - AC² + 2AC² + 2BC² - AB² = 4AD² + 4BE² + 4CF²

⇒ 3AB² + 3AC² + 3BC² = 4AD² + 4BE² + 4CF²

⇒ 3(AB² + AC² + BC²) = 4(AD² + BE² + CF²)

Now,

Given, ∠BAC = 90°

∴ BC² = AB² + AC²

⇒ 3(BC + BC² ) = 4(AD² + BE² + CF²)

⇒ 6BC² = 4(AD² + BE² + CF²)

⇒ 3BC² = 2(AD² + BE² + CF²).

Hope it helps!

Answer 2

math]AE+EB=AB[/math]

2. [math]BD+DC=BC[/math]

3.[math] CF+FA=CA[/math]

adding 1 and 2 gives,

4. [math]AD + DC =  AB + BC[/math]

adding 2 and 3 gives,

5. [math]BF+FA =  BC + CA[/math]

adding 3 and 1 gives,

6. [math]CE+EB= CA + AB[/math]

Adding 4, 5 & 6 gives,

AD+BF+CE + DC+FA+EB = 2(AB+BC+CA)

AD+BF+CE + [math]\frac{1}{2}[/math](AB+BC+CA) = 2(AB+BC+CA)

(since D, E & F are the midpoints of AB, BC and CA)

but, from triangle law of vector addition we have,

[math]AB+BC+CA = 0[/math]

therefore,

[math]AD+BF+CE=0[/math]

Q.E.D

In the above proof I have used the notation