Question
In the figure, given below, $A D \perp B C$.Prove that: $c^2=a^2+b^2-2 a x$.
✓
Answer
Pythagoras theorem states that in a right$-$angled triangle,
the square on the hypotenuse is equal to the sum of the squares on the remaining two sides.
First, we consider the $\triangle ABD$ and applying Pythagoras theorem we get,
$AB^2 = AD^2 + BD^2$
$c^2 = h^2 + ( a - x )^2$
$h^2 = c^2 - ( a - x )^2 ......(i)$
First, we consider the $\triangle ACD$ and applying Pythagoras theorem we get,
$AC^2 = AD^2 + CD^2$
$b^2 = h^2 + x^2$
$h^2 = b^2 - x^2 ......(ii)$
From $(i)$ and $(ii)$ we get,
$c^2 - ( a - x )^2 = b^2 - x^2$
$c^2 - a^2- x^2 + 2ax = b^2 - x^2$
$c^2 = a^2 + b^2 - 2ax$
Hence Proved.
the square on the hypotenuse is equal to the sum of the squares on the remaining two sides.
First, we consider the $\triangle ABD$ and applying Pythagoras theorem we get,
$AB^2 = AD^2 + BD^2$
$c^2 = h^2 + ( a - x )^2$
$h^2 = c^2 - ( a - x )^2 ......(i)$
First, we consider the $\triangle ACD$ and applying Pythagoras theorem we get,
$AC^2 = AD^2 + CD^2$
$b^2 = h^2 + x^2$
$h^2 = b^2 - x^2 ......(ii)$
From $(i)$ and $(ii)$ we get,
$c^2 - ( a - x )^2 = b^2 - x^2$
$c^2 - a^2- x^2 + 2ax = b^2 - x^2$
$c^2 = a^2 + b^2 - 2ax$
Hence Proved.
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