If the square of the length of the longest side of a triangle is equal to the sum of the squares of the lengths of the other two sides, then the triangle is a right triangle.
In the diagram above, if
c^{2 }= a^{2} + b^{2}
then, ΔABC is a right triangle.
Given :
A triangle ABC with the longest side AC and
AB^{2} + BC^{2} = AC^{2}
To Prove :
ΔABC is right a right triangle.
Construction :
Construct a right-angled triangle PQR, right-angled at Q such that PQ = AB and QR = BC.
Proof :
Step 1 :
In ΔPQR, ∠Q = 90°.
Using Pythagorean theorem in ΔPQR, we have
PQ^{2} + QR^{2} = PR^{2} -----(1)
Step 2 :
In ΔABC (given), we have
AB^{2} + BC^{2} = AC^{2} -----(2)
Step 3 :
By construction, PQ = AB and QR = BC.
So, from (1) and (2), we have
PR^{2} = AC^{2}
Get rid of the square from both sides.
PR = AC
Step 4 :
Therefore, by SSS congruence criterion, we get
ΔABC ≅ ΔPQR
which gives
∠B = ∠Q
Step 5 :
But, we have ∠Q = 90° by construction.
Therefore ∠B = 90°.
Hence, ΔABC is a right triangle, right angled at B.
Thus, the theorem is proved.
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