Question -
Answer -
Given, the perpendicular from A on side BC ofa 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, byPythagoras 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]
Therefore, AB2 – AC2 = BC2/2
⇒ 2(AB2 – AC2) = BC2
⇒ 2AB2 – 2AC2 = BC2
∴ 2AB2 = 2AC2 + BC2.