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**In the figure, given below, AD ⊥ BC. **Prove that: c

^{2}= a

^{2}+ b

^{2}- 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,

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 ΔACD 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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