Question

The perpendicular AD on the base BC of a ΔABC intersects BC at D so that DB = 3CD. Prove that 2(AB)^2 = 2(AC)^2 + BC^2.

Answer 1

Given that in ΔABC, we have
AD ⊥BC and BD = 3CD
In right angle triangles ADB and ADC, we have
AB2 = AD2 + BD2 ...(i)
AC2 = AD2 + DC2 ...(ii) [By Pythagoras theorem]
Subtracting equation (ii) from equation (i), we get
AB2 - AC2 = BD2 - DC2
= 9CD2 - CD2 [∴ BD = 3CD] 
= 9CD2 = 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.


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Answer 2

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