Question

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Answer 1

Hope You like My Process
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Formula used is Pythagoras form
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=> (Hypotenuse)² = (base)²+(perpendicular)²
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Given
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=> AB = c

=> BC= a

=> AC = b

=> AD= p

=> AE = h

=> ED = x

=> BD= BC =a/2
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Now for ∆ ADE
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=>(hypotenuse)² = (perpendicular)²+(base)²

=> p² = h² + x²

=> h² = p² - x² ____(1)
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1)

for ∆ AEC
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=>(hypotenuse)² = (perpendicular)²+(base)²

$=> b^2 = h^2 + ( \frac{a}{2} +x)^2 \\ \\ => b^2 = p^2 -x^2 + ( \frac{ {a}^{2} }{4} + 2. \frac{ {}a}{2} .x + {x}^{2} ) \\ \\ = > \boxed{ \bf {b}^{2} = {p}^{2} + ax + \frac{ {a}^{2} }{4} } \: \: \: (proved)$

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2) For ∆ ABE
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=>(hypotenuse)² = (perpendicular)²+(base)²

$= > {c}^{2} = {h}^{2} + ( { \frac{a}{2} - x) }^{2} \\ \\ = > {c}^{2} = {p}^{2} - {x}^{2} + ( \frac{ {a}^{2} }{4} - 2. \frac{a}{2} x + {x}^{2} ) \\ \\ = > \boxed{ \bf{c}^{2} = {p}^{2} - ax + \frac{ {a}^{2} }{4} } \: \: \: (proved)$
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3)

sum of b² and c²
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$\\ = > {b}^{2} + {c}^{2} = ({p}^{2} + ax + \frac{ {a}^{2} }{4} ) + {( {p}^{2} - ax + \frac{ {a}^{2} }{4} )} \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \boxed{ \bf \: 2{p}^{2} + \frac{ {a}^{2} }{2} } \: \: \: \: \: (proved)$
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