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` 

 

Answer 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)] 

 

Answer 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`