Question

In a quadrilateral ABCD, angle A + angle D = 90°. Prove that AC^2 + BD^2 = BC^2 + AD^2​

Answer 1

`AD^(2)+BC^(2)=AE^(2)+ED^(2)+BE^(2)+EC^(2)`

`AC^(2)+BD^(2)=AE^(2)+EC^(2)+BE^(2)+ED^(2)`

Hence prove.

`AC^(2)+BD^(2)=BC^(2)+AD^(2)`

Explanation:

Given:

∠A + ∠D = 90°

We are prove to that

`AC^(2)+BD^(2)=BC^(2)+AD^(2)`

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.

`AD^(2)=AE^(2)+ED^(2)`

`BC^(2)=BE^(2)+EC^(2)`

Now, we add both above equations.

`AD^(2)+BC^(2)=AE^(2)+ED^(2)+BE^(2)+EC^(2)`--------(1)

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

`AC^(2)=AE^(2)+EC^(2)`

`BD^(2)=BE^(2)+ED^(2)`

Now, we add both above equations.

`AC^(2)+BD^(2)=AE^(2)+EC^(2)+BE^(2)+ED^(2)`--------(2)

We get From equation 1 and equation 2.

`AC^(2)+BD^(2)=BC^(2)+AD^(2)`

Hence prove,

`AC^(2)+BD^(2)=BC^(2)+AD^(2)`