Question

1.) Prove that the sum of the squares of the diagonals of a parallelogram is equal to the sum of squares of its sides . Prove properly
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Answer 1

$\large\underline{\sf{Solution-}}$

Let assume that ABCD be the required parallelogram having AC and BD as its diagonals.

Now, we have to prove that

$\rm \: {AC}^{2} + {BD}^{2} = {AB}^{2} + {BC}^{2} + {CD}^{2} + {AD}^{2}$

Construction :- From vertex D, draw DE perpendicular to AB intersecting AB at E and draw CF perpendicular on AB intersecting AB at F when produced.

Now, In triangle AED and triangle BFC

$\rm \: CF = DE \: \{distance \: between \: parallel \: lines \}$

$\rm \: AD = BC \: \{opposite \: sides \}$

$\rm \: \angle \: E \: = \: angle \: F \: \{ \: each \: 90 \degree \: \}$

$\rm\implies \: \triangle \: AED \: \cong \: \triangle \: BFC \: \{ \:RHS \: Rule \}$

$\rm\implies \:AE = BF = x \: (say) \:$

Now, In right angle triangle DEB

Using Pythagoras Theorem, we have

$\rm \: {BD}^{2} = {EB}^{2} + {DE}^{2}$

$\rm \: {BD}^{2} = {(AB - x)}^{2} + {DE}^{2}$

$\rm \: {BD}^{2} = {AB}^{2} + {x}^{2} - 2x \: AB + {DE}^{2}$

$\rm \: {BD}^{2} = {AB}^{2} - 2x \: AB + ({DE}^{2} + {x}^{2})$

$\rm \: {BD}^{2} = {AB}^{2} - 2x \: AB + {AD}^{2} - - - (1)$

Now, In right angled triangle ACF

By using Pythagoras Theorem, we have

$\rm \: {AC}^{2} = {AF}^{2} + {CF}^{2}$

$\rm \: {AC}^{2} = {(AB + x)}^{2} + {CF}^{2}$

$\rm \: {AC}^{2} = {AB}^{2} + {x}^{2} + 2x \: AB + {CF}^{2}$

$\rm \: {AC}^{2} = {AB}^{2} + 2x \: AB + ({CF}^{2} + {x}^{2})$

$\rm \: {AC}^{2} = {AB}^{2} + 2x \: AB + {BC}^{2}$

$\rm \: {AC}^{2} = {CD}^{2} + 2x \: AB + {BC}^{2} - - - - (2)$

$\red{[ \because \: \rm \: ABCD \: is \: a \: parallelogram, \: so \: AB = CD \: ]}$

On adding equation (1) and (2), we get

$\rm \: {AC}^{2} + {BD}^{2} = {AB}^{2} + {BC}^{2} + {CD}^{2} + {AD}^{2}$

Hence, Proved

$\rule{190pt}{2pt}$

Additional Information :-

1. Pythagoras Theorem :-

This theorem states that : In a right-angled triangle, the square of the longest side is equal to sum of the squares of remaining sides.

2. Converse of Pythagoras Theorem :-

This theorem states that : If the square of the longest side is equal to sum of the squares of remaining two sides, angle opposite to longest side is right angle.

3. Area Ratio Theorem :-

This theorem states that :- The ratio of the area of two similar triangles is equal to the ratio of the squares of corresponding sides.

4. Basic Proportionality Theorem

This theorem states that :- If a line is drawn parallel to one side of a triangle, intersects the other two lines in distinct points, then the other two sides are divided in the same ratio.

Answer 2

Given -

• ABCD is a parallelogram in which AC and BD are diagnols intersect at O.

To prove -

• that the sum of the squares of the diagonals of a parallelogram is equal to the sum of squares of its sides. (i.e - AB² + BC² + CD² DA² = AC² + BD²)

Proof -

Since the diagonals of a parallelogram bisect each other. Then, BO and DO are the medians of △ABC and △ADC respectively.

We know that,

the sum of the squares of any two sides of any triangle equals twice the square on half the third side, together with twice the square on the median bisecting the third side (Apollonius's theorem).

Now,

In △ABC (where OB is a median),

$\sf{ {AB}^{2} + {BC}^{2} = 2(OB)^{2} + 2( \frac{1}{2} \: AC )^{2} \:--(i)}$

In △ADC (where OD is a median),

$\sf{ {AD}^{2} + {CD}^{2} = 2(OD)^{2} + 2( \frac{1}{2} \: AC )^{2} \:--(ii)}$

Adding (i) and (ii),

$\small{\sf{ {AB}^{2} + {BC}^{2} + {AD}^{2} + {CD}^{2}= 2(OB)^{2} + 2( \frac{1}{2} \: AC )^{2} + 2(OD)^{2} + 2( \frac{1}{2} \: AC )^{2}}}$

$\small{\sf{ {AB}^{2} + {BC}^{2} + {AD}^{2} + {CD}^{2}= 2(OB)^{2} + \cancel{2} \times \frac{1}{ \cancel{4} ^{2} } \: AC ^{2} + 2(OD)^{2} + \cancel{2} \times \frac{1}{ \cancel{4} ^{2} } \: AC ^{2}}}$

$\small{\sf{ {AB}^{2} + {BC}^{2} + {AD}^{2} + {CD}^{2}= 2 \: OB^{2} + 2 \: OD^{2} + \frac{ {AC}^{2} }{ 2} \: + \frac{AC ^{2}}{ 2} \:}}$

But OB = OD,

$\small{\sf{ {AB}^{2} + {BC}^{2} + {AD}^{2} + {CD}^{2}= 4\:OB^2+AC^2}}$

As we know OB = OD = 1/2 BD,

$\small{\sf{ {AB}^{2} + {BC}^{2} + {AD}^{2} + {CD}^{2}= 4\:(\frac{1}{2}\:BD)^2+AC^2}}$

$\small{\sf{ {AB}^{2} + {BC}^{2} + {AD}^{2} + {CD}^{2}= \cancel{4}\times\frac{1}{\cancel{4}}\:BD^2+AC^2}}$

$\small{\bf{ {AB}^{2} + {BC}^{2} + {AD}^{2} + {CD}^{2}= BD^2+AC^2}}$

Hence Proved

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