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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Solution

In ΔADB, ∠D = $90^{\circ}$ .

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

In ΔADC, ∠D = $90^{\circ}$ .

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

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

= ${A D}^{2}$ + ${D B}^{2}$ + ${B C}^{2}$ + 2DB×BC

= ${A B}^{2}$ + ${B C}^{2}$ + 2BC × BD [Using (i)]

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

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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 [4 MARKS]

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