Get Answers to all your Questions

header-bg qa

Q6   Prove that the sum of the squares of the diagonals of parallelogram is equal to the sum of the squares of its sides.

Answers (1)

best_answer

In parallelogram ABCD, AF and DE are altitudes drawn on DC and produced BA.

In \triangleDEA, by Pythagoras theorem

DA^2=DE^2+EA^2.......................1

In \triangleDEB, by Pythagoras theorem

DB^2=DE^2+EB^2

DB^2=DE^2+(EA+AB)^2

DB^2=DE^2+(EA)^2+(AB)^2+2.EA.AB

DB^2=DA^2+(AB)^2+2.EA.AB....................................2

In \triangleADF, by Pythagoras theorem

DA^2=AF^2+FD^2

In \triangleAFC, by Pythagoras theorem

AC^2=AF^2+FC^2=AF^2+(DC-FD)^2

\Rightarrow AC^2=AF^2+(DC)^2+(FD)^2-2.DC.FD

\Rightarrow AC^2=(AF^2+FD^2)+(DC)^2-2.DC.FD

\Rightarrow AC^2=AD^2+(DC)^2-2.DC.FD.......................3

Since ABCD is a parallelogram.

SO, AB=CD  and BC=AD

In \triangle DEA\, and\, \triangle ADF,

\angle DEA=\angle AFD\, \, \, \, \, \, \, (each 90 \degree)

\angle DAE=\angle ADF      (AE||DF)

AD=AD  (common)

\triangle DEA\, \cong \, \triangle ADF,        (ASA rule)

\Rightarrow EA=DF.......................6

Adding 2 and, we get

DA^2+AB^2+2.EA.AB+AD^2+DC^2-2.DC.FD=DB^2+AC^2\Rightarrow DA^2+AB^2+AD^2+DC^2+2.EA.AB-2.DC.FD=DB^2+AC^2

\Rightarrow BC^2+AB^2+AD^2+2.EA.AB-2.AB.EA=DB^2+AC^2   (From 4 and 6)

\Rightarrow BC^2+AB^2+CD^2=DB^2+AC^2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Posted by

seema garhwal

View full answer