\displaystyle \textbf{Introduction}
\displaystyle \text{Buddhayan, an Indian Mathematician (600 B.C.), developed a relationship}
\displaystyle \text{between the squares described on the hypotenuse of a right-angled triangle}
\displaystyle \text{and the sum of the squares described on the remaining two sides of the}
\displaystyle \text{triangle. But the credit of the present form of this relationship goes to a}
\displaystyle \text{Greek Mathematician Pythagoras and is named as Pythagoras Theorem.}

\displaystyle \textbf{PYTHAGORAS THEOREM}
\displaystyle \textbf{Theorem 9 (Area Based Proof)}
\displaystyle \textit{\text{In a right-angled triangle, the square on the hypotenuse is equal to the sum of the}}
\displaystyle \textit{\text{squares on the remaining two sides.}}
\displaystyle \textbf{Given: }\text{A triangle }ABC\text{ in which }\angle ABC=90^\circ.
\displaystyle \textbf{To Prove: }AC^2=AB^2+BC^2.
\displaystyle \textbf{Construction: Draw squares }ACDE,\ ABFG\text{ and }BCHI
\displaystyle \text{on sides }AC,\ AB\text{ and }BC\text{ respectively. Draw }BMN
\displaystyle \text{perpendicular to }AC\text{ at point }M\text{ and }DE\text{ at point }N.
\displaystyle \text{Join }GC\text{ and }BE.\displaystyle \textbf{Proof:}
\displaystyle \angle FBC=\angle FBA+\angle ABC
\displaystyle \hspace{2.1cm}=90^\circ+90^\circ=180^\circ
\displaystyle \Rightarrow\ FBC\text{ is a straight line.}
\displaystyle \text{Since, }GA\parallel FB\hspace{0.4cm}\text{[Opposite sides of the square]}
\displaystyle \Rightarrow\ GA\text{ is parallel to }FC.
\displaystyle \text{Since, }BMN\text{ and }AE\text{ both are perpendicular to the same line }AC
\displaystyle \Rightarrow\ AE\parallel BMN.
\displaystyle \text{As each angle of quadrilateral }AMNE\text{ is }90^\circ,\ AMNE
\displaystyle \text{is a rectangle.}
\displaystyle \text{Now, let }\angle BAC=x.
\displaystyle \therefore\ \angle GAC=\angle BAE\hspace{1.2cm}\text{[Each }90^\circ+x\text{]}
\displaystyle AG=AB\hspace{3.0cm}\text{[Sides of the same square]}
\displaystyle AC=AE\hspace{3.0cm}\text{[Sides of the same square]}
\displaystyle \therefore\ \triangle GAC\cong\triangle BAE\hspace{1.0cm}\text{[By SAS]}
\displaystyle \Rightarrow\ \text{area of }\triangle GAC=\text{area of }\triangle BAE\qquad\cdots(i)
\displaystyle \text{[Congruent triangles are equal in area]}
\displaystyle \text{We know, the area of a triangle is half the area of a parallelogram}
\displaystyle \text{(rectangle, square, etc.) if both are on the same base and between}
\displaystyle \text{the same parallels.}
\displaystyle \text{Since, }\triangle GAC\text{ and square }ABFG\text{ are on the same base }(AG)
\displaystyle \text{and between the same parallels }(AG\parallel CF)
\displaystyle \therefore\ \text{area of }\triangle GAC=\frac{1}{2}\times\text{area of square }ABFG\qquad\cdots(ii)
\displaystyle \text{Similarly, }\triangle BAE\text{ and rectangle }AMNE\text{ are on the same base }(AE)
\displaystyle \text{and between the same parallels }(AE\parallel BN)
\displaystyle \therefore\ \text{area of }\triangle BAE=\frac{1}{2}\times\text{area of rectangle }AMNE\qquad\cdots(iii)
\displaystyle \text{From equations (i), (ii) and (iii), we get:}
\displaystyle \text{area of square }ABFG=\text{area of rectangle }AMNE\qquad\cdots(iv)
\displaystyle \text{In the same way, it can be proved that:}
\displaystyle \text{area of square }BCHI=\text{area of rectangle }CMND\qquad\cdots(v)
\displaystyle \text{On adding equations (iv) and (v), we get:}
\displaystyle \text{area of square }ABFG+\text{area of square }BCHI
\displaystyle =\text{area of rectangle }AMNE+\text{area of rectangle }CMND
\displaystyle \Rightarrow\ \text{area of square }ABFG+\text{area of square }BCHI
\displaystyle =\text{area of square }ACDE
\displaystyle \Rightarrow\ \text{square on }AB+\text{square on }BC=\text{square on }AC
\displaystyle \Rightarrow\ AB^2+BC^2=AC^2,\ \text{i.e., }AC^2=AB^2+BC^2.
\displaystyle \therefore\ \textbf{Hence Proved.}

\displaystyle \\

\displaystyle \textbf{Converse: }\textit{\text{If in any triangle, the square on the largest side is equal to the}}
\displaystyle \textit{\text{sum of the squares on the remaining two sides, then the triangle is a}}
\displaystyle \textit{\text{right-angled triangle and the angle opposite to the largest side is a right angle.}}
\displaystyle \text{i.e. if in triangle }ABC,\ BC\text{ is the largest side and }BC^2=AB^2+AC^2,
\displaystyle \text{then }\angle A=90^\circ.

\displaystyle \textbf{Theorem 10 (Alternative Proof for Theorem 9)}
\displaystyle \textit{\text{In a right-angled triangle, the square on the hypotenuse is}}
\displaystyle \textit{\text{equal to the sum of the squares of the remaining two sides.}}
\displaystyle \textbf{Given: }\text{A triangle }ABC\text{ in which }\angle ABC=90^\circ.
\displaystyle \textbf{To Prove: }AC^2=AB^2+BC^2.
\displaystyle \textbf{Construction: Draw }BD\perp AC.
\displaystyle \textbf{Proof:}
\displaystyle \textbf{Statement:}\hspace{2.2cm}\textbf{Reason:}
\displaystyle \text{In }\triangle ABC\text{ and }\triangle BDC,
\displaystyle \text{(i) }\angle ABC=\angle BDC\hspace{1.2cm}\text{Each is }90^\circ.
\displaystyle \text{(ii) }\angle BCA=\angle BCD\hspace{1.2cm}\text{Common.}
\displaystyle \therefore\ \triangle ABC\sim\triangle BDC \\ \hspace{0.8cm}\text{A.A. postulate.}
\displaystyle \Rightarrow\ \frac{BC}{DC}=\frac{AC}{BC}\\ \hspace{1.2cm}\text{Corresponding sides of similar}
\displaystyle \hspace{5.8cm}\text{triangles are proportional.}
\displaystyle \Rightarrow\ BC^2=AC\times DC\qquad\cdots I
\displaystyle \text{Now, in }\triangle ABC\text{ and }\triangle ADB,
\displaystyle \text{(i) }\angle ABC=\angle ADB\hspace{1.2cm}\text{Each is }90^\circ.
\displaystyle \text{(ii) }\angle BAC=\angle BAD\hspace{1.2cm}\text{Common.}
\displaystyle \therefore\ \triangle ABC\sim\triangle ADB \\ \hspace{0.8cm}\text{A.A. postulate.}
\displaystyle \Rightarrow\ \frac{AB}{AD}=\frac{AC}{AB} \\ \hspace{1.2cm}\text{Corresponding sides of similar}
\displaystyle \hspace{5.8cm}\text{triangles are proportional.}
\displaystyle \Rightarrow\ AB^2=AC\times AD\qquad\cdots II
\displaystyle \therefore\ AB^2+BC^2=AC\times AD+AC\times DC\hspace{1.0cm}\text{Adding I and II.}
\displaystyle \hspace{2.2cm}=AC(AD+DC)
\displaystyle \hspace{2.2cm}=AC\times AC\hspace{2.4cm}\text{Since }AD+DC=AC.
\displaystyle \hspace{2.2cm}=AC^2
\displaystyle \therefore\ AB^2+BC^2=AC^2.
\displaystyle \therefore\ \textbf{Hence Proved.}

\displaystyle \textbf{Pythagorean Triplets}
\displaystyle \text{Consider three positive numbers }a,\ b\text{ and }c\text{ with }c\text{ as the largest of}
\displaystyle a,\ b\text{ and }c.
\displaystyle \text{If }a^2+b^2=c^2,\ a,\ b\text{ and }c\text{ are called Pythagorean triplets.}
\displaystyle \textbf{For example:}
\displaystyle 3,\ 4\text{ and }5\text{ are Pythagorean triplets as }3^2+4^2=5^2.

\displaystyle \textbf{Application 1: Rhombus}
\displaystyle \text{Since, rhombus is also a parallelogram,}
\displaystyle AB^2+BC^2+CD^2+DA^2=AC^2+BD^2
\displaystyle \Rightarrow\ 4\times(\text{side})^2=AC^2+BD^2.

\displaystyle \textbf{Application 2: Rectangle}
\displaystyle \text{Since, rectangle is also a parallelogram,}
\displaystyle AB^2+BC^2+CD^2+DA^2=AC^2+BD^2
\displaystyle \Rightarrow\ l^2+b^2+l^2+b^2=d^2+d^2
\displaystyle \hspace{2.5cm}\left[\because\ AC=BD=d=\text{diagonal}\right]
\displaystyle \Rightarrow\ 2l^2+2b^2=2d^2
\displaystyle \Rightarrow\ l^2+b^2=d^2.

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