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Let AD be a median of the Delta ABC. If AE and AF are medians of the triangle ABD and ADC, respectively, and AD = m_(1), AE = m_(2), AF = m_(3), " then " a^(2)//8 is equal to |
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Answer» `m_(2)^(2) + m_(3)^(2) - 2m_(1)^(2)`  In `DeltaABC, AD^(2) = m_(1)^(2) = (c^(2) + b^(2))/(2) -(a^(2))/(4)` In `DeltaABD, AE^(2) = m_(2)^(2) = (AD^(2) + c^(2))/(2) - (((a)/(2))^(2))/(4)` [Apolloniun THEOREM] In `DeltaADC, AF^(2) = m_(3)^(2) = (AD^(2) + b^(2))/(2) -(((a)/(2))^(2))/(4)` `:. m_(2)^(2) + m_(3)^(2) = AD^(2) + (b^(2) + c^(2))/(2) - (a^(2))/(8)` `= m_(1)^(2) + m_(1)^(2) + (a^(2))/(4) -(a^(2))/(8) = 2m_(1)^(2) + (a^(2))/(8)` or `m_(2)^(2) + m_(3)^(2) - 2m_(1)^(2) = (a^(2))/(8)` |
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