Question

plz solve Q.14 .... no spamming plz ​

Answer 1

Answer:

✌hope it helps u✌️

Step-by-step explanation:

Given:

In ΔABC, we have

AD ⊥BC

BD = 3CD

In right angle triangles ADB and ADC, we have

AB²= AD² + BD²...(i)

AC² = AD² + DC² ...(ii) [By Pythagoras theorem]

Subtracting equation (ii) from equation (i), we get

AB² - AC² = BD² - DC²

= 9CD² - CD² [∴ BD = 3CD]

= 9CD² = 8(BC/4)² [Since, BC = DB + CD = 3CD + CD = 4CD]

Therefore, AB² - AC² = BC²/2

⇒ 2(AB² - AC²) = BC²

⇒ 2AB² - 2AC² = BC²

∴ 2AB² = 2AC² + BC².

Answer 2

Given :

AD ⊥ BC

BD = 3CD

To Prove :

2AB² = 2AC² + BC²

Proof :

we have BD = 3CD

$\therefore$ BC = BD + CD

BC = 3CD + CD

BC = 4CD

CD = 1/4 BC ⠀⠀⠀⠀⠀⠀⠀......(1)

In ∆ADB, ∠ADB = 90°

so using pythagoras theorem,

AB² = AD² + BD² ⠀⠀⠀......(2)

Similarly,

In ∆ADC, ∠ADC = 90°

using pythagoras theorem,

AC² = AD² + CD² ⠀⠀⠀......(3)

On subtracting (3) and (2), we get

AB² - AC² = AD² + BD² + AD² + CD²

AB² - AC² = BD² + CD²

AB² - AC² = [(3CD)² - (CD²)] (∵ BD = 3CD)

AB² - AC² = 9CD² - CD²

AB² - AC² = 8CD²

AB² - AC² = 8 × (1/4 BC)² ⠀⠀[using (1) ]

AB² - AC² = 8 × 1/16 BC²

AB² - AC² = 1/2BC²

2(AB² - AC²) = BC²

2AB² - 2AC² = BC²

Hence proved !