Question

13. In A ABC, seg AD I seg BC
DB = 3CD. Prove that:
2ABP = 2AC2 + BC2 solve and show me​

Answer 1

Given:

• ∆ABC
• AD⊥BC
• DB = 3CD

To Prove:

$\sf{2AB² = 2AC² + BC²}$

Solution:

it is given that,

$\sf{DB = 3CB}$

$\sf{∴ BC = 4CD......(1)}$

According to Pythagoras theorem,

in ∆ABD,

$\sf{AB² = AD² + DB²}$

$\sf{⟹AD² = AB² + DB²........(2)}$

in ∆ACD,

$\sf{AC² = AD² + CD²}$

$\sf{⟹AC² = (AB² - DB²) + CD² (from (2))}$

$\sf{⟹AC² = AB² - (3CD)² + CD² (given)}$

$\sf{⟹AC² = AB² - 9CD² + CD²}$

$\sf{⟹AC² = AB² - 8CD²}$

$\sf{⟹AB² = AC² + 8CD²}$

$\sf{⟹\:AB² = AC² + 8(\frac{BC}{4})²\:\:\:(from (1))}$

$\sf{⟹\:AB² = AC² + \frac{BC²}{2}}$

$\sf{⟹\: 2AB² = 2AC² + BC²}$

Hence,

$\sf\underline\color{pink}{2AB² = 2AC² + BC²}$

Proved ✔