Question

if in triangle ABC AD is median and AM is perpendicular to BC the prove that AC2=AD2+BC*DM+1/4 BC2

Answer 1

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 :)

Answer 2

here is your answer ☺☺☺☺☺☺☺☺


D is the mid point of BC and AE⊥BC.
In right ∆ ABE,

AB'2 = AE'2 + BE'2 … (1) (Pythagoras Theorem)
In right ∆ ADE,

AD'2 = AE2 + ED2 … (2) (Pythagoras Theorem)

From 1 and 2, we got.............................
From (1) and (2), we get

AB² = AD² - MD² + BM²

⇒ AB² = AD² - DM² + (BD - DM)²

⇒ AB² = AD² - DM² + BD² + DM² - 2BD × DM

⇒ AB² = AD² - 2BD × DM + BD²

⇒ AB² = AD² - 2(BC/2) × DM + (BC/2)²   {∵ BD = DC = BC/2}

⇒ AB² = AD² - BC × DM + BC²/4