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,
AB2 = AD2 + BD2 …………….(i)
AC2 = AD2 + DC2 ………………..(ii)
Subtracting equation (ii) from equation (i), we get
AB2 – AC2 = BD2 – DC2
= 9CD2 – CD2 [Since, BD = 3CD]
= 8CD2
= 8(BC/4)2 [Since, BC = DB + CD = 3CD + CD = 4CD]
AB2 – AC2 = BC2/2
⇒ 2(AB2 – AC2) = BC2
⇒ 2AB2 – 2AC2 = BC2
∴ 2AB2 = 2AC2 + BC2.
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