Question

In a quadrilateral ABCD, ∠B = 90° and ∠D = 90°. Prove that :
2AC² - AB² = BC² + CD² + DA²

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Answer 1

Given :-

In a quadrilateral ABCD, ∠B = 90° and ∠D = 90°.

To Prove :-

2AC² - AB² = BC² + CD² + DA²

Solution :-

In △ABC , using pythagoras theorem

➞ AC² = AB² + BC²( i )

In △ADC, using pythagoras theorem

➞AC² = AD² + DC²__ ( ii )

Adding (i) and (ii)

AC² + AC ² = AB ² + BC ² + AD ² + DC ²

2AC ² = AB ² + BC ² + AD + DC

2AC ² - AB² = BC ² + CD ² +DA²

Hence Proved.

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Answer 2

Given

• ABCD is a quadrilateral
• ∠B = 90°
• ∠D = 90°

To find

• 2 AC² - AB² = BC² + CD²+ DA²

Solution

In ∆ABC

By pythagoras theorem,

AC² = AB² + BC² ---- (1)

In ∆ADC

By pythagoras theorem,

AC² = CD² + DA² --- (2)

Add (1) and (2),

AC² + AC² = AB² + BC² + CD² + DA²

2AC² = AB² + BC² + CD² + DA²

2AC² - AB² = BC² + CD² + DA²

Hence, proved.

Figure :-

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Know More :-

Pythagoras theorem :-

In a right - angled triangle, the sum of the square of the hypotenuse is equal to the sum of the square of the other two sides.

• H² = P² + B²

where,

• H = hypotenuse (the longest side)
• P = perpendicular
• B = base

Converse of pythagoras theorem :-

If the square of the length of the longest side of a triangle is equal to the sum of the squares of the other two sides, then the triangle is a right angled triangle.