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.

Asked by Pragya Singh | 1 year ago |  65

#### 1 Answer

##### Solution :-

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($$\dfrac{BC}{4}$$)[Since, BC = DB + CD = 3CD + CD = 4CD]

Therefore, AB2 – AC2 = BC2/2

⇒ 2(AB2 – AC2) = BC2

⇒ 2AB2 – 2AC2 = BC2

2AB2 = 2AC2 + BC2.

Answered by Abhisek | 1 year ago

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