Given that ABCD be a parallelogram.
Construction : We have to Draw AF and DE perpendicular on side DC and on extended side AB to use Pythagoras theorem
In ΔDEA, DE2 + EA2 = DA2 … (i)
Similarly, In ΔDEB,
DE2 + EB2 = DB2
DE2 + (EA + AB)2 = DB2 ( EB =EA + AB)
DE2 + EA2 + AB2 + 2EA × AB == DB2
(DE2 + EA2) + AB2 + 2EA × AB = DB2
DA2 + AB2 + 2EA × AB = DB2 … (ii)( Using -----1)
In ΔAFC, AC2 = AF2 + FC2
= AF2 + (DC − FD)2
= AF2 + DC2 + FD2 − 2DC × FD
= (AF2 + FD2) + DC2 − 2DC × FD
=AC2 = AD2 + DC2 − 2DC × FD ( using AD2 = AF2 + FD2)… (iii)
Now,given that ABCD is a parallelogram, AB = CD … (iv) And, BC = AD … (v)
Look In ΔDEA and ΔADF,
∠DEA = ∠AFD (Both 90°)
∠EAD = ∠ADF (EA || DF -Alternate angles )
AD = AD (Common)
∴ ΔEAD cong. ΔFDA (AAS congruence criterion)
⇒ EA = DF … (vi)
Adding equations (i) and (iii), we obtain
DA2 + AB2 + 2EA × AB + AD2 + DC2 − 2DC × FD = DB2 + AC2
DA2 + AB2 + AD2 + DC2 + 2EA × AB − 2DC × FD = DB2 + AC2
BC2 + AB2 + AD2 + DC2 + 2EA × AB − 2AB × EA = DB2 + AC2
[Using equations (iv) and (vi)]
AB2 + BC2 + CD2 + DA2 = AC2 + BD2
Q. In Fig. is the perpendicular bisector of the line segment DE, FA perpendiculat OB and F E intersects OB at the point C. Prove that 1/ OA + 1/OB = 1/OC