Question

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Answer 1

Given :-

In a ABCD square ,BD = ED.

The bisector of ∠ABD intersect the line CD at point E.

C - D - E

Required To Prove:-

BE² = 2 ( 2 + √2 ) AB²

Proof :-

Given that

□ ABCD is a square.

We know that

All sides are equal in a square.

AB = BC = CD = DA ---------------(1)

BD is the diagonal of the square ABCD

We know that

The diagonal of a square = √2 × side

BD =√2 AB =√2 BC =√2 CD =√2 DA -------(2)

Now, In ∆ BEC , a right angled triangle

By Pythagoras Theorem,

BE² = BC² + CE²

From the given figure,

CE = CD+DE

Therefore, BE² = BC²+(CD+DE)²

From (1)

=> BE² = AB² +(AB+DE)²--------------(3)

In ∆ BCD , a right angled triangle

By Pythagoras Theorem,

BD² = BC²+CD²

=> BD² = AB²+AB² ---(From (1))

=> BD² = 2 AB²

Given that

BD = ED

Therefore, ED² = 2AB²

=> ED = √(2AB)²

=> ED = √2 AB -----------(4)

On substituting the value of ED in (3) then

BE² = AB² + (AB+√2 AB)²

=>BE²=AB²+[AB²+2(AB)(√2 AB)+(√2 AB)²]

Since, (a+b)² = a²+2ab+b²

Where, a = AB and b = √2 AB

=> BE² = AB² +(AB²+2√2 AB²+2 AB²)

=> BE² = AB²+AB²+2AB²+2√2 AB²

=> BE² = 4AB² +2√2 AB²

On taking 2 AB² as common factor then

=> BE² = 2AB² ( 2 + √2)

Therefore, BE² = 2 ( 2 + √2 ) AB²

Hence, Proved.

Used Formulae :-

♦ The diagonal of a square = √2 × side

♦ (a+b)² = a²+2ab+b²

♦ All sides are equal in a square.

Used Theorem:-

Pythagoras Theorem:-

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