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8. In a AABC, ADL BC and BD-3CD. Prove that 2AB2AC+ BC |
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Answer» Given that in ΔABC, we haveAD ⊥BC and BD = 3CDIn right angle triangles ADB and ADC, we haveAB°2 = AD^2 + BD^2 ...(i)AC^2 = AD^2 + DC^2 ...(ii) [By Pythagoras theorem]Subtracting equation (ii) from equation (i), we getAB^2 - AC^2 = BD^2 - DC^2= 9CD^2 - CD^2 [∴ BD = 3CD] = 9CD2 = 8(BC/4)^2 [Since, BC = DB + CD = 3CD + CD = 4CD]Therefore, AB^2 - AC^2 = BC^2/2⇒ 2(AB^2 - AC^2) = BC^2⇒ 2AB^2 - 2AC^2 = BC^2∴ 2AB^2 = 2AC^2 + BC^2 thanks |
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