Question

the perpendicular from a on side BC of a triangle ABC intersect BC at D such that BD is equal to 3 CD prove that 2 a b square is equal to 2 AC square + BC square​

Answer 1

Question : -

A perpendicular from A on side BC of a △ABC intersects BC at D such that BD = 3 CD. Prove that, 2AB² = 2 AC² + BC².

Answer : -

Given that,

∠ADC = ∠ADB = 90° (as BD ⊥ BC)

BD = 3 CD

To prove,

2AB² = 2AC² + BC²

Assumption

Let BD = 3y, CD = y, and thus, by adding BD and CD we get, BC = 4y. As BC is a straight line.

Proof

In right - angled △ADC and △ADB,

we have,

[By using Pythagoras theorem],

AB² = AD² + BD² (AB is the hypotenuse side of △ADB) ..... (i)

AC² = AD² + DC² (AC is the hypotenuse side of △ADC) ..... (ii)

Now we'll subtract equation (ii) from equation (i).

By subtracting equation (ii) from equation (i), we get,

AB² - AC² = (AD² - AD²) + (BD² - DC²)

AB² - AC² = $\cancel{AD^{2} - AD^{2}}$ + BD² - DC²

AB² - AC² = BD² - CD²

AB² - BD² = AC² - CD²

By substituting the assumed values,

AB² - (3y)² = AC² - y²

AB² - 9y² = AC² - y²

AB² = AC² + 9y² - y²

AB² = AC² + 8y²

[By multiplying all the terms with 2],

2AB² = 2AC² + 16y²

Which equals to,

2AB² = 2AC² + BC²

[As BC = 4y, ∴ BC² = 16y²]

Hence, the proof.

Answer 2

Answer:

Follow the steps given above.