The perpendicular AD on base BC of $△$ ABC intersects BC at D, so that BD=3CD.
Prove that 2AB² =2AC² +BC².

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Solution

$In △ ABC, the perpendicular AD on base BC intersects BC at D, such that BD = 3 CD .$

$△ ABD is a right - angled triangle . On using Pythagoras' Theorem, we get : {AB}^{2} = {AD}^{2} + {BD}^{2} . . . (1) So, △ ACD is a right - angled triangle . Thus, on using Pythagoras' T heorem, we get : {AC}^{2} = {AD}^{2} + {CD}^{2} . . . (2) By (1) - (2), we get : A B^{2} - A C^{2} = A D^{2} + B D^{2} - A D^{2} - C D^{2} \Rightarrow A B^{2} - A C^{2} = B D^{2} - C D^{2}$

$On multiplying both sides by 2, we get : 2 ({AB}^{2} - {AC}^{2}) = 2 ({BD}^{2} - {CD}^{2}) \Rightarrow 2 {AB}^{2} - 2 {AC}^{2} = 2 ({BD}^{2} - {CD}^{2}) \Rightarrow 2 {AB}^{2} = 2 {AC}^{2} + 2 ({BD}^{2} - {CD}^{2}) On using (a^{2} - b^{2}) = (a + b) (a - b), we get : 2 {AB}^{2} = 2 {AC}^{2} + 2 (BD + CD) (BD - CD) \Rightarrow 2 {AB}^{2} = 2 {AC}^{2} + 2 BC (3 CD - CD) [BD + CD = BC and BD = 3 CD] \Rightarrow 2 {AB}^{2} = 2 {AC}^{2} + 2 BC \times 2 CD \Rightarrow 2 {AB}^{2} = 2 {AC}^{2} + BC \times 4 CD \Rightarrow 2 {AB}^{2} = 2 {AC}^{2} + BC \times (3 CD + CD) \Rightarrow 2 {AB}^{2} = 2 {AC}^{2} + BC \times (BD + CD) [BD = 3 CD] \Rightarrow 2 {AB}^{2} = 2 {AC}^{2} + BC \times BC [BD + CD = BC] \Rightarrow 2 AB = 2 {AC}^{2} + {BC}^{2}$

Hence proved.

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