Pythagoras Theorem

Pythagoras Theorem:

In a right triangle, the square on the hypotenuse is equal to the sum of the squares on the other two sides.

Pythagoras Theorem Diagram - Pythagoras Theorem
Given: A right-angle triangle ABC in which
∠B = 90°

To prove: AC2 = AB2 + BC2

Construction: From B draw BD ⊥ AC

Proof: In triangles ADB and ABC, we have
∠ADB = ∠ABC [Each equal to 90°]
and ∠A = ∠A [common]

So, by AA-similarity, we have
△ADB ~ △ABC
⇒ AD/AB = AB/AC [∵ corresponding sides of similar triangles are proportional]
⇒ AB2 = AD x AC ……….(i)

In △s BDC and ABC, we have ∠CDB = ∠ABC [Each equal to 90°]
and ∠C = ∠C [common]

So by AA- similarity, we have
△BDC ~ △ABC
⇒ DC/BC = BC/AC [∵ corresponding sides of similar triangles are proportional]
⇒ BC2 = AC x DC ……….(ii)

Adding (i) and (ii), we get AB2 + BC2 = AD x AC + AC x DC
⇒ AB2 + BC2 = AC (AD + DC)
⇒ AB2 + BC2 = AC x AC
⇒ AB2 + BC2 = AC2
Hence, AC2 = AB2 + BC2

Converse of the Pythagoras Theorem:

In a triangle, if the square on one side is equal the sum of the squares on the remaining two, the angle opposite to first side is a right angle.

Converse of the Pythagoras Theorem - Pythagoras Theorem
Given: △ABC such that AC2 = AB2 + BC2

To prove: ∠ABC = 90°

Construction: Construct △PQR such that PQ = AB, QR = BC and ∠PQR = 90°

Proof: In △PQR, ∠PQR = 90° [By construction]
∴ PR2 = PQ2 + QR2 [By Pythagoras Theorem]
or PR2 = AB2 + BC2 [∵ PQ = AB, QR = BC, by construction]
But AC2 = AB2 + BC2 [given]
∴ PR2 = AC2 ⇒ PR = AC

Now in △ABC and △PQR,
AB = PQ [By const.]
BC = QR [By const.]
AC = PR [Proved above]
∴ △ABC ≅ △PQR [SSS congruency theorem]
Then ∠B = ∠Q [Correspoding parts of congruent triangles]
But ∠Q = 90° [By constructive]]
Hence ∠B = 90° [Proved]

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