For class 10th Triangles chapter how i solve the problem from exercise 6.5 of question no.14, give me the best way to solve the problem of this question The perpendicular from A on side BC of a Δ ABC intersects BC at D such that DB = 3CD (see Figure). Prove that 2AB2 = 2AC2 + BC2.
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]
Therefore, AB2 – AC2 = BC2/2
⇒ 2(AB2 – AC2) = BC2
⇒ 2AB2 – 2AC2 = BC2
∴ 2AB2 = 2AC2 + BC2.