Question

In figure ,ad is a median and am perpendicular to bc Ad²+bc.dm+(bc/2)² equals to

Answer 1

Answer:

Solution:-

D is the midpoint of BC and AM ⊥ BC.

In right angled triangle ABM,

AB² = AM² + BM² ....(1) - Pythagoras Theorem

In right angled triangle ADM,

AD² = AM² + MD² ....(2) - Pythagoras Theorem

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

Hence proved.