Mathematics
ABC is a triangle, right-angled at B. M is a point on BC. Prove that :
AM2 + BC2 = AC2 + BM2.
Pythagoras Theorem
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Answer
By formula,
By pythagoras theorem,
⇒ (Hypotenuse)2 = (Perpendicular)2 + (Base)2
In right-angled △ ABM,
⇒ AM2 = AB2 + BM2
⇒ AB2 = AM2 - BM2 ………..(1)
In right-angled △ ABC,
⇒ AC2 = AB2 + BC2
⇒ AB2 = AC2 - BC2 ………..(2)
From equations (1) and (2), we get :
⇒ AM2 - BM2 = AC2 - BC2
⇒ AM2 + BC2 = AC2 + BM2.
Hence, proved AM2 + BC2 = AC2 + BM2.
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