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**ABC is a triangle, right-angled at B. M is a point on BC.**

Prove that: AM^{2} + BC^{2} = AC^{2} + BM^{2}.

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#### Solution

The pictorial form of the given problem is as follows,

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 ΔABM and applying Pythagoras theorem we get,

AM^{2 }= AB + BM^{2}

AB^{2 }= AM^{2} - BM^{2 } ......(i)

Now, we consider the ΔABC and applying Pythagoras theorem we get,

AC^{2 }= AB^{2} + BC^{2}

AB^{2 }= AC^{2} - BC^{2 } ......(ii)

From (i) and (ii) we get,

AM^{2 }- BM^{2} = AC^{2} - BC^{2}

AM^{2 }+ BC^{2 }= AC^{2} + BM^{2 }

Hence Proved.^{}

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