The perpendicular from A on side BC of a Δ ABC intersects BC at D such that DB = 3CD
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06/06/2021 12:16 pm
Given, the perpendicular from A on side BC of a Δ ABC intersects BC at D such that;
DB = 3CD.
In Δ ABC,
AD ⊥BC and BD = 3CD
In right angle triangle, ADB and ADC, by Pythagoras theorem,
AB^{2} =^{ }AD^{2} + BD^{2} …………….(i)
AC^{2} =^{ }AD^{2} + DC^{2} ………………..(ii)
Subtracting equation (ii) from equation (i), we get
AB^{2} – AC^{2} = BD^{2} – DC^{2}
= 9CD^{2} – CD^{2} [Since, BD = 3CD]
= 8CD^{2}
= 8(BC/4)^{2 }[Since, BC = DB + CD = 3CD + CD = 4CD]
AB^{2} – AC^{2} = BC^{2}/2
⇒ 2(AB^{2} – AC^{2}) = BC^{2}
⇒ 2AB^{2} – 2AC^{2} = BC^{2}
∴ 2AB^{2} = 2AC^{2} + BC^{2}.
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