Question

prove that the sum of the squares of the sides of a parallelogram is equal to the sum of the squares of its diagonals

Answer 1

In parallelogram ABCD, AB = CD, BC = AD
Draw perpendiculars from C and D on AB as shown.
In right angled ΔAEC, AC = AE + CE [By Pythagoras theorem]
⇒ AC = (AB + BE) + CE
⇒ AC = AB + BE + 2 AB × BE + CE → (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)


Hope it is right
ΔAFD ≅ ΔBEC (RHS congruence rule)
⇒ AF = BE
Consider right angled ΔDFB
BD = BF + DF [By Pythagoras theorem]
= (EF – BE) + CE [Since DF = CE]
= (AB – BE) + CE [Since EF = AB]
⇒ BD = AB + BE – 2 AB × BE + CE → (2)
Add (1) and (2), we get
AC + BD = (AB + BE + 2 AB × BE + CE ) + (AB + BE – 2 AB × BE + CE )
= 2AB + 2BE + 2CE
AC + BD = 2AB + 2(BE + CE ) → (3)
From right angled ΔBEC, BC = BE + CE [By Pythagoras theorem]
Hence equation (3) becomes,
AC + BD = 2AB + 2BC
= AB + AB + BC + BC
= AB + CD + BC + AD
AC + BD = AB + BC + CD + AD
Thus the sum of the squares of the diagonals of a parallelogram is equal to the sum of the squares of its sides.

Hope it works

Answer 2

Answer:

Step-by-step explanation:

⚡ HEY MATE HERE IS YOUR ANSWER ⚡

⚡HOPE ❄️IT ❄️HELP❄️ YOU ⚡

⭐☺️THANKS ⭐