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