Question

Prove that in any parallelogram, the sum of the squares of all sides is equal to the sum of the squares of the diagonals.

DON'T SPAM

IT'S URGENT. . . ​

Answer 1

Answer:

i.e. O is the mid point of AC and BD. In ∆ABD, point O is the midpoint of side BD. In ∆CBD, point O is the midpoint of side BD. Hence, the sum of the squares of the diagonals of a parallelogram is equal to the sum of the squares of its sides

Step-by-step explanation:

please mark my ans as brainliest please

Answer 2

✧══════•❁❀❁•══════✧

$\large \underline \mathtt \fcolorbox{green}{blue}{\color{pink}{★彡⛄αηsωєя⛄彡★}}$

•❁❀❁•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 + 2AB × 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 – 2AB × BE + CE² → (2)

•❁❀❁•Add (1) and (2),

we get AC² + BD² = (AB² + BE² + 2AB × BE + CE²)+ (AB² + BE² – 2AB × 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.

✧══════•❁❀❁•══════✧

✧══════•❁❀❁•══════✧

$\large\boxed{\fcolorbox{Blue}{cyan}{hope\:it\:helps\: you}}$

$\large\boxed{\fcolorbox{green}{blue}{\color{yellow}{MARK\:ME\:AS\: BRAINLIST}}}$

✧══════•❁❀❁•══════✧. . . .