Abc is a Triangle, Right-angled at B. M is a Point on Bc Prove That: Am2 + Bc2 = Ac2 + Bm2 - Mathematics

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Sum

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 = AB  + 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,
AM- BM2  = AC2  - BC2 
AM+ BC= AC2 + BM2  
Hence Proved.

  Is there an error in this question or solution?
Chapter 13: Pythagoras Theorem [Proof and Simple Applications with Converse] - Exercise 13 (B) [Page 163]

APPEARS IN

Selina Concise Mathematics Class 9 ICSE
Chapter 13 Pythagoras Theorem [Proof and Simple Applications with Converse]
Exercise 13 (B) | Q 3 | Page 163

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