Question

ans this sum..............​

Answer 1

Answer:

Step-by-step explanation:

Given:

ABC is a Δ

AD ⊥ CB

DB = 3CD

To prove:

2AB² = 2AC² + BC²

Poof:

AB² = AD² + DB²

AD² = AB² - DB² ........(i)

AC² = AD² + CD²

⇒AD² = AC² - CD².......................(ii)

Comparing (i) and (ii)

AB² - DB² = AC² - CD²

AB² - (3CD)² = AC² - CD² ........(iii)

BC= CD+DB

BC= CD+3CD

BC=4CD

CD=BC/4........(iv)

Substituting (iv) in (iii)

AB² - (3BC/4)² = AC² - (BC/4)²

⇒AB² - 9BC²/16 = AC² - BC²/16

⇒(16AB² - 9BC²)/16 = (16AC² - BC²)/16

16 will get cancel

we will get,

16AB² - 9BC² = 16AC² - BC²

⇒16AB² - 16AC² = -BC² + 9BC²

⇒ 16AB² = 16AC² + 8BC²

⇒ 8 × 2AB² = 8(2AC² + BC²)

⇒ 2AB² = 2AC² + BC²

Answer 2

Answer:

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