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Question
In the figure, given below, AD ⊥ BC.
Prove that: c2 = a2 + b2 - 2ax.
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Solution
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 ΔABD and applying Pythagoras theorem we get,
AB2 = AD2 + BD2
c2 = h2 + ( a - x )2
h2 = c2 - ( a - x )2 ......(i)
First, we consider the ΔACD and applying Pythagoras theorem we get,
AC2 = AD2 + CD2
b2 = h2 + x2
h2 = b2 - x2 ......(ii)
From (i) and (ii) we get,
c2 - ( a - x )2 = b2 - x2
c2 - a2 - x2 + 2ax = b2 - x2
c2 = a2 + b2 - 2ax
Hence Proved.
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