Prove that in a parallelogram sum of the squares of the sides is equal to sum of the squares of the diagonals.
Answers
Hi dear friend..
See the attached file.. There is figure..
SOLUTION is here
In parallelogram ABCD, AB = CD, BC = AD
Draw perpendiculars from C and D on AB as shown.
In right angled ΔAEC, AC2 = AE2 + CE2 [By Pythagoras theorem]
⇒ AC2 = (AB + BE)2 + CE2
⇒ AC2 = AB2 + BE2 + 2 AB × BE + CE2 → (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)
ΔAFD ≅ ΔBEC (RHS congruence rule)
⇒ AF = BE
Consider right angled ΔDFB
BD2 = BF2 + DF2 [By Pythagoras theorem]
= (EF – BE)2 + CE2 [Since DF = CE]
= (AB – BE)2 + CE2 [Since EF = AB]
⇒ BD2 = AB2 + BE2 – 2 AB × BE + CE2 → (2)
Add (1) and (2), we get
AC2 + BD2 = (AB2 + BE2 + 2 AB × BE + CE2) + (AB2 + BE2 – 2 AB × BE + CE2)
= 2AB2 + 2BE2 + 2CE2
AC2 + BD2 = 2AB2 + 2(BE2 + CE2) → (3)
From right angled ΔBEC, BC2 = BE2 + CE2 [By Pythagoras theorem]
Hence equation (3) becomes,
AC2 + BD2 = 2AB2 + 2BC2
= AB2 + AB2 + BC2 + BC2
= AB2 + CD2 + BC2 + AD2
∴ AC2 + BD2 = AB2 + BC2 + CD2 + AD2
Thus the sum of the squares of the diagonals of a parallelogram is equal to the sum of the squares of its sides.
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Answer:
Step-by-step explanation:
In parallelogram ABCD, AB = CD, BC = AD
Draw perpendiculars from C and D on AB as shown.
In right angled ΔAEC, AC2 = AE2 + CE2 [By Pythagoras theorem]
⇒ AC2 = (AB + BE)2 + CE2
⇒ AC2 = AB2 + BE2 + 2 AB × BE + CE2 → (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)
ΔAFD ≅ ΔBEC (RHS congruence rule)
⇒ AF = BE
Consider right angled ΔDFB
BD2 = BF2 + DF2 [By Pythagoras theorem]
= (EF – BE)2 + CE2 [Since DF = CE]
= (AB – BE)2 + CE2 [Since EF = AB]
⇒ BD2 = AB2 + BE2 – 2 AB × BE + CE2 → (2)
Add (1) and (2), we get
AC2 + BD2 = (AB2 + BE2 + 2 AB × BE + CE2) + (AB2 + BE2 – 2 AB × BE + CE2)
= 2AB2 + 2BE2 + 2CE2
AC2 + BD2 = 2AB2 + 2(BE2 + CE2) → (3)
From right angled ΔBEC, BC2 = BE2 + CE2 [By Pythagoras theorem]
Hence equation (3) becomes,
AC2 + BD2 = 2AB2 + 2BC2
= AB2 + AB2 + BC2 + BC2
= AB2 + CD2 + BC2 + AD2
∴ AC2 + BD2 = AB2 + BC2 + CD2 + AD2
Thus the sum of the squares of the diagonals of a parallelogram is equal to the sum of the squares of its sides.