Question

The prependicular from A on the side BC of a angle ABC intersects BC at D such that DB=3CD. Prove that:2AB^2=2AC^2+2BC^2​

Answer 1

$\huge\boxed{\mathfrak{\fcolorbox{red}{orange}{HOLA\:MATE!}}}$

$<b>Hey! There is mistake in your question. We have to prove: 2AB^2=2AC^2 + BC^2.$

$<b><u>Answer$

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.

Answer 2

Answer

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.