Answer:
[tex]AD^{2}+BC^{2}=AE^{2}+ED^{2}+BE^{2}+EC^{2}[/tex]
[tex]AC^{2}+BD^{2}=AE^{2}+EC^{2}+BE^{2}+ED^{2}[/tex]
Hence prove.
[tex]AC^{2}+BD^{2}=BC^{2}+AD^{2}[/tex]
Step-by-step explanation:
Given:
∠A + ∠D = 90°
We are prove to that
[tex]AC^{2}+BD^{2}=BC^{2}+AD^{2}[/tex]
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.
[tex]AD^{2}=AE^{2}+ED^{2}[/tex]
[tex]BC^{2}=BE^{2}+EC^{2}[/tex]
Now, we add both above equations.
[tex]AD^{2}+BC^{2}=AE^{2}+ED^{2}+BE^{2}+EC^{2}[/tex]--------(1)
Similarly, Using Pythagoras Theorem for triangle AEC and triangle BED.
[tex]AC^{2}=AE^{2}+EC^{2}[/tex]
[tex]BD^{2}=BE^{2}+ED^{2}[/tex]
Now, we add both above equations.
[tex]AC^{2}+BD^{2}=AE^{2}+EC^{2}+BE^{2}+ED^{2}[/tex]--------(2)
We get From equation 1 and equation 2.
[tex]AC^{2}+BD^{2}=BC^{2}+AD^{2}[/tex]
Hence prove,
[tex]AC^{2}+BD^{2}=BC^{2}+AD^{2}[/tex]
