Question

AD is the median of a triangle ABC and AM is perpendicular to BC. Prove that AC²+AB²=2AD²+BC²/2​

Answer 1

Answer:

Mark as the brainliest brother please.

Step-by-step explanation:

We have AD as the median and AM perpendicular to BC

Since, AD is median, D is the midpoint of BC. hence,

BD = CD

Now, since AM is perpendicular to BC, so AMC is a right angled triangle. Hence,

AC² = AM² + MC²

Now, MC² can be written as (DM + DC)²

=> AC² = AM² + (DM + DC)²

=> AC² = AM² + DM² + DC² + 2DC × DM

(since, (a + b)² = a² + b² + 2ab)

Now, 2DC = BC (since D is the midpoint of BC)

=> AC² = AM² + DM² + DC² + BC × DM

Now, AM² = AD² - DM² (By Pythagoras Theorem, in ∆AMD)

=> AC² = AD² - DM² + DM² + DC² + BC × DM

=> AC² = AD² + BC × DM + DC²

Now, DC = 1/2 BC

=> DC² = (1/2 × BC)²

=> DC² = 1/4 × BC²

So put DC² as 1/4 BC² and we have

AC² = AD² + BC × DM + 1/4 × BC²

Hence Proved :)