Question

In a right Angeled Triangle right angled at A, D is any point on AB. Then prove that

DC² = BC² + BD² -2AB.BD

Answer 1

As the given triangle is a tight angled triangle.



By Pythagoras Theorem,

In ∆ABC,
BC² = AB² + AC² ----: ( 1 )


In ∆ADC,
= > DC² = ( AD )² + AC²
= > DC² = ( AB - BD )² + AC²
= > DC² = AB² + BD² - 2AB.BD + AC² ----: ( 2 )



Now, subtract ( 1 ) from ( 2 ),

= > DC² - BC² = AB² + BD² - 2AB.BD + AC² - AB² - AC²

= > DC² - BC² = BD² - 2AB.BD

= > DC² = BD² + BC² - 2AB.BD.


Hence, proved.

$\:$

Answer 2

$\huge\textbf\green{Answer :}$


$\textbf{Given :}$ A triangle in which right angled at A.
D is a point on AB.

$\textbf{To Prove :}$ DC² = BC² + BD² - 2AB.BD.

$\textbf{Proof :}$

In ∆ABC

By Pythagoras theorem...

BC² = AC² + AB²

AC² = BC² - AB² ...... (1)

In ∆ADC

DC² = AD² + AC²

We can write AD as AB - BD.

So,

DC² = (AB - BD)² + AC²

DC² = AB² + BD² - 2AB.BD + AC² ...... (2)

Put value of (1) in (2)

Then,

DC² = AB² + BD² - 2AB.BD + BC² - AB²

DC² = BC² + BD² - 2AB.BD

Hence, Proved...

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