In a quadrilateral abcd, angle a + angle d = 90°. prove that ac^2 + bd^2 = bc^2 + ad^2

Hence prove.

Step-by-step explanation:

Given:

∠A + ∠D = 90°

We are prove to that

Solution:

See required figure in attached file.

We know sum of the all angles of a triangle is 180°.

So, in triangle AED.

∠A + ∠E + ∠D = 180°

∠E + (∠A + ∠D) = 180°

Now, we substitute ∠A + ∠D = 90° in above equation.

∠E + 90° = 180°

∠E = 180° - 90°

∠E = 90°

Using Pythagoras Theorem for triangle ADE and triangle BEC.

Now, we add both above equations.

--------(1)

Similarly, Using Pythagoras Theorem for triangle AEC and triangle BED.

Now, we add both above equations.

--------(2)

We get From equation 1 and equation 2.

Hence prove,

the simplified form is

step-by-step explanation:

we want to simplify the expression,

let us change the middle bar to a normal division sign to obtain,

we multiply the first fraction by the reciprocal of the second fraction to obtain,

we factor to obtain,

we cancel the common factors to get,

we simplify to get,

the correct answer is b

5+2x=a

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