Question
$\text{ABC}$ 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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Answer
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 $\triangle A B M$ and applying Pythagoras theorem we get,
$AM ^2= AB + BM ^2$
$A B^2=A M^2-B M^2 \dots......(i)$
Now, we consider the $\triangle A B C$ and applying Pythagoras theorem we get,
$A C^2=A B^2+B C^2$
$A B^2=A C^2-B C^2\dots......(ii)$
From $(i)$ and $(ii)$ we get,
$A M^2 \cdot B M^2=A C^2-B C^2$
$A M^2+B C^2=A C^2+B M^2$
Hence Proved
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 $\triangle A B M$ and applying Pythagoras theorem we get,
$AM ^2= AB + BM ^2$
$A B^2=A M^2-B M^2 \dots......(i)$
Now, we consider the $\triangle A B C$ and applying Pythagoras theorem we get,
$A C^2=A B^2+B C^2$
$A B^2=A C^2-B C^2\dots......(ii)$
From $(i)$ and $(ii)$ we get,
$A M^2 \cdot B M^2=A C^2-B C^2$
$A M^2+B C^2=A C^2+B M^2$
Hence Proved
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