## Pythagoras Theorem:

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

Given: A right-angle triangle ABC in which ∠B = 90° To prove: AC^{2} = AB^{2} + BC^{2}Construction: From B draw BD ⊥ ACProof: 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] ⇒ AB ^{2} = 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] ⇒ BC ^{2} = AC x DC ……….(ii)Adding (i) and (ii), we get AB ^{2} + BC^{2} = AD x AC + AC x DC⇒ AB ^{2} + BC^{2} = AC (AD + DC)⇒ AB ^{2} + BC^{2} = AC x AC⇒ AB ^{2} + BC^{2} = AC^{2}Hence, AC ^{2} = AB^{2} + BC^{2} |

## 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.

Given: △ABC such that AC^{2} = AB^{2} + BC^{2}To prove: ∠ABC = 90°Construction: Construct △PQR such that PQ = AB, QR = BC and ∠PQR = 90°Proof: In △PQR, ∠PQR = 90° [By construction]∴ PR ^{2} = PQ^{2} + QR^{2 }[By Pythagoras Theorem]or PR ^{2} = AB^{2} + BC^{2} [∵ PQ = AB, QR = BC, by construction]But AC ^{2} = AB^{2} + BC^{2} [given]∴ PR ^{2} = AC^{2} ⇒ PR = ACNow 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] |