Method 1: By similarity:
Prerequisites:
Similarity of triangles (definition)
AA similarity (proof)
Proof:
.png)
Let \triangle ABC be a right triangle, right angled at B. Let us draw an altitude from vertex B to the line AC, meeting AC at a point D.
Now, in \triangle ADB and \triangle ABC,
\qquad\quad\:\!\angle A = \angle A\qquad\qquad\qquad\qquad\quad\:\text{(common)}\\ \qquad\quad\angle ADB = \angle ABC = 90^o\\ \therefore\quad\;\;\;\triangle ADB\sim\triangle ABC\qquad\qquad\qquad\text{(by AA similarity rule)}
Since, by definition of similar triangles, corresponding sides of similar triangles are proportional,
\therefore\quad\;\;\dfrac{AD}{AB} = \dfrac{AB}{AC}\\[12pt] \Rightarrow\quad\;\; AB^2 = AD \times AC\qquad\qquad\qquad\cdots\text{(1)}
Similarly, from \triangle BDC and \triangle ABC,
\qquad\quad BC^2 = DC \times AC\qquad\qquad\qquad\cdots\text{(2)}
Adding (1) and (2),
\qquad\quad\begin{align} AB^2 + BC^2 &= AD\times AC + DC\times AC\\ &= (AD + DC)\times AC\\ &= AC\times AC\\ &= AC^2\end{align}
Q.E.D.
Method 2: By Area:
Prerequisites:
Square (definition)
Area of square (proof)
Area of right triangle (proof)
SAS congruence (proof)
Angle sum property of triangle (proof)
Angle on a straight line (proof)
(a+b)^2 = a^2 + b^2 (proof)
Proof:
.png)
Let ABCD be a square having side length equal to (a+b). Let us locate the points E, F, G and H on the lines AB, BC, CD and DA respectively, such that AE = BF = CG = DH = a. Join EF, FG, GH and HE by straight lines, as shown in the figure.
Let \angle AEH = \theta, then by angle sum property of a triangle,
\qquad\quad\begin{align}\angle AHE &= 180^o - \angle A - \angle AEH\\ &= 180^o - 90^o - \theta\\ &= 90 -\theta\qquad\qquad\qquad\qquad\cdots\text{(1)}\end{align}
Now, in \triangle HAE and \triangle EBF,
\qquad\quad AE = BF = a\qquad\qquad\qquad\text{(by construction)}\\ \qquad\quad AH = BE = b\qquad\qquad\qquad\text{(by construction)}\\ \qquad\quad\angle A = \angle B = 90^o\qquad\qquad\quad\:\!\text{(by definition of square)}
Hence, \triangle HAE\cong\triangle EBF by SAS rule.
Thus, by CPCTC,
\qquad\quad\angle BEF = \angle AHE = 90^o - \theta\qquad\qquad\qquad\text{(from $(1)$)}\\ \begin{align}\therefore\quad\;\;\;\angle HEF &= 180^o - \angle AEH - \angle BEF\qquad\quad\text{($\because$ AB is a straight line)}\\ &= 180^o - \theta - (90^o - \theta)\\ &= 90^o\qquad\qquad\qquad\qquad\qquad\qquad\;\cdots\text{(2)}\end{align}
Similarly, all the four triangles, i.e., \triangle HAE,\;\triangle EBF,\;\triangle FCG and \triangle GDH are congruent. Thus, by CPCTC, \angle HEF, \; \angle EFG,\; \angle FGH and \angle GHE are equal and, from (2), are equal to 90^o. Also, EF = FG = GH = HE = c (say). Hence, EFGH is a square (see definition of a square).
Further, since these four triangles are congruent, they have same area.
Now, let us find the area of the square ABCD:
\qquad\quad \text{Area of }\; ABCD = \text{Area of } \;EFGH + ar (\triangle HAE) + ar (\triangle EBF)+ ar (\triangle FCG) \\ \\ \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad + ar (\triangle GDH)\\ \Rightarrow\quad\;\; (a+b)^2 = c^2 + 4\times\frac{1}{2} ab\qquad\qquad\;\text{(by formula for area of square and right triangle)}\\ \Rightarrow\quad\;\; a^2 + b^2 + 2ab = c^2 + 2ab\qquad\qquad\text{($\because\;\;(a+b)^2 = a^2 + b^2 + 2ab$)}\\ \Rightarrow\quad\;\; a^2 + b^2 = c^2\qquad\qquad\qquad\qquad\quad\;\;\cdots\text{(3)}
Since, a and b are chosen arbitrarily, hence the result of equation (3) holds for all right triangles.
Q.E.D.
No comments:
Post a Comment