the perpendicular from A on BC of a triqngle ABC intersects BC at D such that DB =3CD. prove that 2AB² =2AC²+bc²?
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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.
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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