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Statement of Pythagoras' Theorem

ABC is an obt...

Question

$A B C$ is an obtuse angle triangle & $A D ⊥ B C$ produced. Process that ${A C}^{2} + {A B}^{2} + {B C}^{2} + 2 B C - B D$

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Solution

Let us consider the problem:

In $Δ A D C$

$A C^{2} = A D^{2} + D C^{2}$ (Pythagoras theorem

$A C^{2} = A D^{2} + {(D B + B C)}^{2}$

$A C^{2} = A D^{2} + {(D B + B C)}^{2}$

$A C^{2} = A B^{2} + B C^{2} + 2 D B \times B C .$

$[∴ A D^{2} + D B^{2} = A B^{2}]$

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In a ΔABC, obtuse angled at B, if AD is perpendicular to CB produced, prove that AC = ${A B}^{2}$ + ${B C}^{2}$ + 2BC $\times$ BD

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