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Question
ABC is a triangle, right-angled at B. M is a point on BC.
Prove that: AM2 + BC2 = AC2 + BM2
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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,
AM2 = AB2 + BM2
AB2 = AM2 - BM2 ...(i)
Now, we consider the ΔABC and applying Pythagoras theorem we get,
AC2 = AB2 + BC2
AB2 = AC2 - BC2 ...(ii)
From (i) and (ii) we get,
AM2 - BM2 = AC2 - BC2
AM2 + BC2 = AC2 + BM2
Hence, Proved.
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