Question

Class 10 . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Maths . . . . . . . . . . . . . . . . . . . . . . . . . . . .
PLs answer fast . . . . . . . . . . . . . . . . . . . .
If in a triangle ABC , AD is median and AM is perpendicular to BC, then prove that
AC^2 = AD ^2 + BC * DM + 1/4 BC^2

Answer 1

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.

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