P and Q are the mid-points of the sides CA and CB respectively of a `triangleABC` , right angled at C, prove that. (i) `4AQ^(2)=4AC^(2)+BC^(2)` (ii) `4BP^(2)=4BC^(2)+AC^(2)`

Explanation

(i) `triangleAQC , angleC= 90^(@)`
Using Pythagoras theorem
`AQ^(2)=AC^(2)+QC^(2)`
`AQ^(2)= AC^(2) + ((BC)/2)^(2)`
(Q is the mid-point of BC)
`AQ^(2)= AC^(2)+(BC^(2))/4`
`4AQ^(2)= 4AC^(2)+BC^(2)` …(1) Hence proved.
(ii) `triangleBPC, angleC=90^(@)` ,
Using Pythagoras theorem, we have
`BP^(2)=BC^(2) + PC^(2)` ,
`BP^(2)=BC^(2)+ ((AC)/2)^(2)` (since P is the mid-point of AC)
`Rightarrow " " 4 BP^(2)= 4BC^(2)+AC^(2)` ...(2) Hence proved.
(iii) Adding (1) and (2) we get
`4AQ^(2)+4BP^(2)=4AC^(2)+BC^(2)+4BC^(2)+AC^(2)`
`4(AQ^(2)+BP^(2))= 5(AC^(2)+BC^(2))`
`Rightarrow " " 4(AQ^(2)+BP^(2))= 5AB^(2) " " ( AC^(2) +BC^(2)=AB^(2))` Hence proved.