Question

in triangle ABC , angle b is 90 degrees and D is the mid point of BC . Prove that AC²=AD²+3CD²​

Answer 1

Given:-

• ∠B of triangle ABC is 90°.
• D is mid point of side BC of triangle ABC.

To prove:-

• AC² = AD²+ 3CD²

SolutiOn:-

If ∠B is 90° and D is mid point. So, one right angle will be ∆ABC and second Right angle will be ∆ABD.

We know that,

Pythagoras theorem is

◆ (Hypotenuse)² = (Perpendicular)² + (Base)² ◆

So,

In ∆ABC

⇒ AC² = AB² + BC² (Pythagoras theorem)

⇒ AC² - BC² = AB²

Or,

⇒AB² = AC² - BC² --------------(i)

In ∆ABD

⇒ AD² = AB² + BD²

⇒ AD² - BD² = AB²

Or,

⇒AB² = AD² - BD² --------------(ii)

From equation (i) and (ii)

⇒ AC² - BC² = AD² - BD²

(Beacuse both are equal to AB²)

⇒ AC² = AD² - BD² + BC²

⇒ AC² = AD - BD² + (2CD)²

(∵ BC = BD + DC and BD and DC are equal beacuse D is mid point of BC)

⇒ AC² = AD² - CD² + (2CD)²

(BD = CD beacuse D is mid point of BC)

⇒ AC² = AD² - 3CD²

Hence, Proved

Therefore,

AC² = AD²+ 3CD²

Answer 2

Answer:

Step-by-step explanation: