In a triangle, if the square of one side is equal to the sum of the squares of the other two sides, then prove that the angle opposite the first side is a right angle.
In ABC, the square of the one side is equal to the sum of the square of the other two sides. AB2 + BC2 = AC2.
To Prove:
Construction: we construct a PQR, right-angled at Q such that PQ = AB and QR = BC
Proof: Now from PQR
PR2 = PQ2 + QR2 [Pythagorus Theorem as ]
or, PR2 = AB2 + BC2 [By construction] -(i)
But, AC2 = AB2 + BC2 [Given] -(ii)
So, AC2 = PR2 [From (i) and (ii)]
AC = PR -(iii)
Now in ABC and PQR
AB = PQ [By construction]
BC = QR [By construction]
AC = PR [Proved in (iii)]
So, ABC PQR
Therefore
[CPCT]
But [By Construction]
so, , hence, proved.