In a sight angled A the square
of hypotenuse is equal to the
Sum of squares of remaining two
sides.
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1
Answer:
Given: A right-angled triangle ABC, right-angled at B.
To Prove- AC2 = AB2 + BC2
Construction: Draw a perpendicular BD meeting AC at D.
Pythagoras Theorem Proof
Proof:
We know, △ADB ~ △ABC
Therefore, ADAB=ABAC (corresponding sides of similar triangles)
Or, AB2 = AD × AC ……………………………..……..(1)
Also, △BDC ~△ABC
Therefore, CDBC=BCAC (corresponding sides of similar triangles)
Or, BC2= CD × AC ……………………………………..(2)
Adding the equations (1) and (2) we get,
AB2 + BC2 = AD × AC + CD × AC
AB2 + BC2 = AC (AD + CD)
Since, AD + CD = AC
Therefore, AC2 = AB2 + BC2
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