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then prove that AC2 = AD2 + BC Ă DM +d BC247. If in ABC, AD is median and AMä¸BC, |
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Answer» Solution:-D is the midpointof BC and AM⊥ BC.In right angled triangleABM,AB² = AM² + BM² ....(1) - Pythagoras TheoremIn right angled triangle ADM,AD² = AM² + MD² ....(2) - Pythagoras TheoremFrom (1)and(2), we getAB² = 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 + BC²/4Hence proved. |
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