Verify the identity geometrically: $(a + b + c)^{2} = a^{2} + b^{2} + c^{2} + 2 a b + 2 b c + 2 a c$

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Solution

Step 1: Draw a square and cut into $9$ parts.
Step 2: There are $3$ squares (red, yellow, green) and $6$ rectangles ( $2$ pink, $2$ purple, $2$ blue)
Step 3: Area of the full square $=$ $(a + b + c)^{2}$
Step 4: Now we have to find the area of $3$ inside square(red, yellow, green) = $a^{2} + b^{2} + c^{2}$
Step 5: Consider the area of $2$ pink rectangle $=$ length $\times$ breadth $= b . a + b . a = 2 a b$
Step 6: Area of $2$ purple rectangle $= a . c + a . c = 2 a c$ and Area of $2$ blue rectangle $= b . c + b . c = 2 b c$
Step 7: Area of full square $=$ area of $3$ inside square $+$ area of $2$ pink rectangle $+$ area of $2$ purple rectangle $+$ area of $2$ blue rectangle.
i.e., $(a + b + c)^{2} = a^{2} + b^{2} + c^{2} + 2 a b + 2 b c + 2 a c$
Hence, geometrically we proved the identity $(a + b + c)^{2} = a^{2} + b^{2} + c^{2} + 2 a b + 2 b c + 2 a c$

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Δ A B C

c o s A = \frac{b^{2} + c^{2} - a^{2}}{2 b c}, c o s B = \frac{c^{2} + a^{2} - b^{2}}{2 a c}, c o s C = \frac{a^{2} + b^{2} - c^{2}}{2 a b}

If a + b + c = 2s, then prove the following identities

(a) s² + (s − a)² + (s − b)² + (s − c)² = a² + b² + c²

(b) a² + b² − c² + 2ab = 4s (s − c)

(d) a² − b² − c² + 2ab = 4(s − b) (s − c)

(e) (2bc + a² − b² − c²) (2bc − a² + b² + c²) = 16s (s − a) (s − b) (s − c)

(f)

(a + b + c)^{2} = a^{2} + b^{2} + c^{2} + 2 a b + 2 b c + 2 c a

a = 2

b = 4

c = 6

a = 5

b = - 2

c = 3

a = - 2

b = 3

c = - 4

\frac{a^{2} - b^{2} - 2 b c - c^{2}}{a^{2} + b^{2} + 2 a b - c^{2}}

△ A B C

cos A = \frac{b^{2} + c^{2} - a^{2}}{2 b c}

cos B = \frac{c^{2} + a^{2} - b^{2}}{2 c a}, cos C = \frac{a^{2} + b^{2} - c^{2}}{2 a b}

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