In the given figure, $Δ A B C$ is an obtuse triangle, obtuse angled at B. If $A D ⊥ C B$ , then prove that ${A C}^{2} = {A B}^{2} + {B C}^{2} + 2 B C . B D$

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Solution

Given: $A B C$ is obtuse triangle.
${A B}^{2} = {A D}^{2} + {D B}^{2} ⟶ (1)$
${A C}^{2} = {A D}^{2} + {D C}^{2} ⟶ (2)$

from $(1)$ ${A D}^{2} = {A B}^{2} - {D B}^{2}$

$∴ {A C}^{2} = {A B}^{2} - {D B}^{2} + {(D B + B C)}^{2} [∵ B C = D B + B C]$

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

${A C}^{2} = {A B}^{2} + {B C}^{2} + 2 B C \times D B$ $(P r o v e d)$

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In fig., $A B C$ is a triangle in which $∠ A B C > 90^{o}$ and $A D ⊥ C B$ produced. Prove that $A C^{2} = A B^{2} + B C^{2} + 2 B C . B D$ .

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