Question

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

Answer 1

Answer:

$AD^{2}+BC^{2}=AE^{2}+ED^{2}+BE^{2}+EC^{2}$

$AC^{2}+BD^{2}=AE^{2}+EC^{2}+BE^{2}+ED^{2}$

Hence prove.

$AC^{2}+BD^{2}=BC^{2}+AD^{2}$

Step-by-step 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,

$AC^{2}+BD^{2}=BC^{2}+AD^{2}$