Q33.Prove that the sum of the squares of the diagonals of parallelogram is equal to
the sum of the squares of its sides.
Answers
In parallelogram ABCD,
AB=CD and BC=AD
Draw perpendiculars from C and D on AB as shown.
In right angled △AEC,
⇒ (AC)² = (AE)² + (CE)²
[ By Pythagoras theorem ]
⇒ (AC)² = (AB+BE)² + (CE)²
⇒ (AC)² = (AB)² + (BE)² + 2×AB×BE+(CE)²
----- ( 1 )
From the figure CD=EF [ Since CDFE is a rectangle ]
But CD=AB
⇒ AB=CD=EF
Also CE=DF [Distance between two parallel lines]
⇒ △AFD≅△BEC [ RHS congruence rule ]
⇒ AF=BE [ CPCT ]
Considering right angled △DFB
⇒ (BD)² = (BF)² + (DF)² [ By Pythagoras theorem ]
⇒ (BD)² = (EF−BE)² + (CE)² [ Since DF=CE ]
⇒ (BD)² =(AB−BE)² +(CE)² [ Since EF=AB ]
⇒ (BD)² =(AB)² + (BE)² −2×AB×BE+(CE)² ----- ( 2 )
Adding ( 1 ) and ( 2 ), we get
(AC)² + (BD)2 =(AB)² + (BE)² + 2×AB×BE + (CE)²
(AB)² +(BE)² −2×AB×BE+(CE)²
⇒ (AC)² +(BD)² = 2(AB)² +2(BE)² +2(CE)²
⇒ (AC)² +(BD)² = 2(AB)² +2[(BE)² +(CE)² ] ---- ( 3 )
In right angled △BEC,
⇒ (BC)² = (BE)² +(CE)² [ By Pythagoras theorem ]
Hence equation ( 3 ) becomes,
⇒ (AC)² +(BD)² = 2(AB)² + 2(BC)²
⇒ (AC)² +(BD)² =(AB)² +(AB)² +(BC)² +(BC)²
∴ (AC)² +(BD) =(AB)² +(BC) +(CD)² +(AD)²
∴ The sum of the squares of the diagonals of a parallelogram is equal to the sum of squares of its sides.