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 [4 MARKS]

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Solution

Concept : 1 Mark
Application : 1 Mark
Proof : 2 Marks

In $Δ A D B, ∠ D = 90^{\circ}$ .

$∴ {A D}^{2} + {D B}^{2} = {A B}^{2} . . . . . . . . . (1)$ [ By Pythagoras Theorem]

In $Δ A D C, ∠ D = 90^{\circ}$ .

$∴ {A C}^{2} = {A D}^{2} + {D C}^{2}$ [ By Pythagoras Theorem]

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

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

$\Rightarrow {A C}^{2} = {A B}^{2} + {B C}^{2} + 2 B C \times B D$ [Using (1)]

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

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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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