Question

In ∆ ABC, seg AD perpendicular seg BC

DB = 3CD. Prove that ः

2AB^2

= 2AC^2

+ BC^2​

Answer 1

Answer:

hope it's help u......

Answer 2

$\bf{\underline{Given}}:-$

• ABC is a triangle in which AD ⊥ BC and,

• BD = 3CD

To prove :-

• 2AB² = 2AC² + BC²

Proof :-

we have,

BD = 3CD

therefore,

BC = (BD + CE) = 4CD

Now,

⤇ CD = ¼ BC .................................(i)

Then,

In ∆ ADB,

• ∠ ADB = 90°

By Pythagoras theorem,

AB² = AD² + BD² .........................(ii)

again,

In ∆ ADC,

• ∠ ADC = 90°

By Pythagoras theorem,

AC² = AD² + CD² .......................(iii)

Subtracting eq (iii) from (ii)

(AB²) - (AC)² = (AD² + BD²) - (AD² + CD²)

⤇ (AB² - AC²) = (BD² - CD²)

⤇ [(3CD)² - (CD)²]

⤇ 8CD² ⠀ ⠀ ⠀⠀ ⠀ ⠀ ⠀ ⠀ [ ∴ BD = 3CD ]

By using equation (i),

⤇ 8 × (1/16 BC²)

⤇ ½ BC²

∴ 2AB² - 2AC² = BC²

Hence, 2AB² = (2AC² + BC²)

Proved ....... :)