Question

ABCD is a square, BD=ED. Bisector of angle ABD intersects line CD in point E, and C-D-E, then show that BE2=2(2+root2) AB2

Answer 1

Answer:

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Answer 2

Given:

ABCD is a square

BD = ED

Bisector of ∠ABD intersects line CD in point E, and C, D, E

To Find:

BE² = 2(2+√2)AB²

Solution:

It is given that ABCD is a square.

By definition of the square, all the sides of a square are equal. This means that,

AB = BC = CD = AD  ..(i)

Also, each angle of a square is 90°, which implies

∠A = ∠B = ∠C = ∠D = 90°

Then, as given BD is the diagonal of the square. So, we know that diagonal of the square is √2 times that of the side of the square.

∴ BD = √3AB ..(ii)

It is also given that,

BD = DE

ED = √2AB [from(ii)]  ..(iii)

Now, in ΔBDE,

∠BDE > 90°

ΔBDE  is an obtuse-angled triangle.

So, as ∠BCD that is ∠BCE is 90°.

BC⊥CD⊥CE [C-D-E]

Thus, ΔBDE also contains an altitude BC in its exterior.

Hence, an obtuse-angled triangle has perpendicular property, we get,

BE² = BD²+DE²+2CD×DE

now, substituting the measures,

       = (√2AB)²+(√2AB)²+2AB√2AB  [using (i),(ii) and (iii)]

Simplifying the values

       = (2AB² + 2AB²)+2√2AB²

       = (2AB² + 2AB²) + 2√2AB²

       = 4AB² + 2√2AB²

       = 2AB²(2+√2)

       = 2(2+√2)AB²

Therefore, BE² = 2(2+√2)AB².