Question

ad perpendicular to bc prove ab^2+cd^2=bd^2+ac^2

Answer 1

In ΔABC as AD⊥BC
Then ∠ADB = ∠ADC ⇒ 90°

In ΔADB          ∠ADB ⇒ 90°
By Pythagoras theorem,
AB² = BD² + AD²
AD² = AB² - BD²                    ....(1)

Also
In ΔADC          ∠ADC ⇒ 90°
By Pythagoras theorem,
AC² = CD² + AD²
AD² = AC² - CD²                    ....(2)

From (1) and (2)
AD² ⇒ AB² - BD² = AC² - CD²    
           AB² + CD² = BD² + AC²
                                                                                           HENCE PROVED


:) Hope this Helps!!!

Answer 2

Angle ABC and CBS are right angles
So by Pythagoras theorem
(1) AB ^2+BC^2=AC^2
or, BC^2=AC^2-AB^2......(1)
(2) BC^2+BD^2=CD^2
or, BC^2=CD^2-BD^2........(2)


From (1)&(2)

AC^2-AB^2=CD^2-BD^2
AC^2+BD^2=CD^2+AB^2