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Question
In a ABC , AD is a median and AL ⊥ BC .
Prove that
(a) `AC^2=AD^2+BC DL+((BC)/2)^2`
(b) `AB^2=AD^2-BC DL+((BC)/2)^2`
(c) `AC^2+AB^2=2.AD^2+1/2BC^2`
Solution 1
(a) In right triangle ALD
Using Pythagoras theorem, we have
`=AC^2-AL^2+LC^2`
`=AD^2-DL^2+(DL+DC)^2 ` [Using (1)]
=`AD^2-DL^2+(DL+(BC)/2)^2` [∵ AD is a median]
`=AD^2-DL^2+DL^2+((BC)/2)^2+BC.DL `
∴` AC^2=AD^2+BC.DL+((BC)/2)^2` .................(2)
(b) In right triangle ALD
Using Pythagoras theorem, we have
`AL^2=AD^2-DL^2` ....................(3)
Again, In right triangle ABL
Using Pythagoras theorem, we have
`AB^2=AL^2+LB^2`
`=AD^2-DL^2+LB^2` [𝑈𝑠𝑖𝑛𝑔 (3)]
Solution 2
(a) In right triangle ALD
Using Pythagoras theorem, we have
`=AC^2-AL^2+LC^2`
`=AD^2-DL^2+(DL+DC)^2 ` [Using (1)]
=`AD^2-DL^2+(DL+(BC)/2)^2` [∵ AD is a median]
`=AD^2-DL^2+DL^2+((BC)/2)^2+BC.DL `
∴` AC^2=AD^2+BC.DL+((BC)/2)^2` .................(2)
(b) In right triangle ALD
Using Pythagoras theorem, we have
`AL^2=AD^2-DL^2` ....................(3)
Again, In right triangle ABL
Using Pythagoras theorem, we have
`AB^2=AL^2+LB^2`
`=AD^2-DL^2+LB^2` [𝑈𝑠𝑖𝑛𝑔 (3)]
=`AD^2-DL^2+(BD-DL)^2`
=`AD^2DL^2+(1/2BC-DL)^2`
=`AD^2-DL^2+((BC)/2)^2-BC.DL+DL^2`
∴ AB^2=AD^2-BC.DL+((BC)/2)^2 .............(4)
(c) Adding (2) and (4), we get,
=`AC^2+AB^2=AD^2+BC.DL+((BC)/2)^2+AD^2-BC.DL+((BC)/2)^2`
=`2AD^2+(BC^2)/4+(BC^2)/4`
=`2AD^2+(BC^2)/4+(BC^2)/4`
=`2AD^2+1/2BC^2`
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