Question

If in triangle ABC, AD is a median and AM is perpendicular to BC, then prove that AB2 = AD2-BC*DM + 1/4 BC2. 1. See answer. Add answer ... D is the midpoint of BC and AM ⊥ BC. In right angled triangle ABM, ... DM + BC²/4. Hence proved. tramwayniceix and 418 more users found this answer helpfu

Answer 1

Answer:

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Answer 2

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.