Question

The perpendicular from A on side BC of a
triangle ABC intersects BC at D such that DC= 3 DB . Prove that 2AB²=2AC² + BC².

Answer 1

$\bf \to \: \green{ \underline{given \div }}$

=> perpendicular from A on sides BC of a

=> ∆ABC intersect BC at D

=> DC = 3DB

$\bf \to \: \orange{ \underline{to \: prove \div }} \\ \\ \rm \to \: 2ab {}^{2} = 2ac {}^{2} + bc {}^{2}$

$\bf \to \: { \underline{solution \div }}$

$\rm \to \: in \: \triangle \: acd \: \\ \\ \rm \to \: by \: apply \: pythagorus \: theorm \\ \\ \rm \to \: {h}^{2} = {p}^{2} + {b}^{2} \\ \\ \rm \to \: {ac}^{2} = {ad}^{2} + {cd}^{2} \\ \\ \rm \to \: {ad}^{2} = {ac}^{2} - {cd}^{2} ....(1) \\ \\ \rm \to \: in \triangle \: abd \\ \\ \rm \to \: {ab}^{2} = {ad}^{2} + {db}^{2} \\ \\ \rm \to \: {ad}^{2} = {ab}^{2} - {db}^{2} .....(2) \\ \\ \rm \to \: from \: equation \: (1) \: and \: (2) \\ \\ \rm \to \: {ac}^{2} - {cd}^{2} = {ab}^{2} - {db}^{2}.....(3) \\ \\ \rm \to \: 3dc \: = db \\ \\ \rm \to \: let \: bc \: = x \\ \\ \rm \to \: cd \: + \: db \: = x \\ \\ \rm \to \: cd + 3cd = x \\ \\ \rm \to \: cd \: = \frac{x}{4} \\ \\ \rm \to \: cd \: = \frac{bc}{4} \\ \\ \rm \to \:put \: the \: value \: in \: equation \: (3) \\ \\ \rm \to \: (ac) {}^{2} - ( \frac{bc}{4} ) {}^{2} = (ab) {}^{2} - (3dc) {}^{2} \\ \\ \rm \to \: (ac) {}^{2} - \frac{(bc) {}^{2} }{16} = (ab) {}^{2} - 3 (\frac{bc}{4} ) {}^{2} \\ \\ \rm \to \: (ac) {}^{2} - \frac{(bc) {}^{2} }{16} = (ab) {}^{2} - \frac{9bc {}^{2} }{16} \\ \\ \rm \to \: (ac) {}^{2} - (ab) {}^{2} = - \frac{9bc {}^{2} }{16} + \frac{(bc) {}^{2} }{16} \\ \\ \rm \to \: (ac) {}^{2} - (ab) {}^{2} = \frac{bc {}^{2} }{2} \\ \\ \rm \to \: 2ab {}^{2} = 2ac {}^{2} + bc {}^{2}$

Answer 2

$\tt GIVEN:-$

$\tt \triangle \: ABC \: is \: a \: triangle\\ \tt AD \perp BC \\ \tt D C = 3DB$

$\tt PROVE:-$

$\tt 2AB {}^{2} = 2 AC {}^{2} + BC {}^{2}$

$\tt SOLUTION:-$

$\tt in \triangle \: ADB \: and \: \triangle ADC$

$\tt AB {}^{2} = AD {}^{2} + BD {}^{2} ....(i) \: \: \: \: \: \: (by \: pythogoras \: theorem) \\ \tt AC {}^{2} = AD {}^{2} + DC {}^{2} ......(ii)$

$\tt sub \: (ii) \: from \: (i)$

$\tt AB {}^{2} - AC {}^{2} = BD {}^{2} - DC {}^{2}$

$\tt 9CD {}^{2} - CD {}^{2} \: \: ( \therefore DC = 3DB )$

$\tt 9CD {}^{2} = (\frac{BC}{4} {)}^{2}$

$\tt \therefore AB {}^{2} -AC {}^{2} = \frac{ {BC}^{2} }{2}$

$\tt = > 2( AB {}^{2} -AC {}^{2} )= {BC}^{2}$

$\tt = > 2 AB {}^{2} - 2AC {}^{2} = {BC}^{2}$

$\tt = > 2 AB {}^{2} = 2AC {}^{2} + {BC}^{2}$