The perpendicular from A on the side BC of a triangle intersects BC AT D such that DB=3CD. Prove that 2AB^2=2AC^2+BC^2.
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Given that in ΔABC, we have
AD ⊥BC and BD = 3CD
In right angle triangles ADB and ADC, we have
AB^2 = AD^2 + BD^2 ...(i)
AC^2 = AD^2 + DC^2 ...(ii) [By Pythagoras theorem]
Subtracting equation (ii) from equation (i), we get
AB^2 - AC^2 = BD^2 - DC^2
= 9CD^2 - CD^2 [∴ BD = 3CD]
= 9CD^2 = 8(BC/4)^2 [Since, BC = DB + CD = 3CD + CD = 4CD]
Therefore, 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.
AD ⊥BC and BD = 3CD
In right angle triangles ADB and ADC, we have
AB^2 = AD^2 + BD^2 ...(i)
AC^2 = AD^2 + DC^2 ...(ii) [By Pythagoras theorem]
Subtracting equation (ii) from equation (i), we get
AB^2 - AC^2 = BD^2 - DC^2
= 9CD^2 - CD^2 [∴ BD = 3CD]
= 9CD^2 = 8(BC/4)^2 [Since, BC = DB + CD = 3CD + CD = 4CD]
Therefore, 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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