For class 10th Triangles chapter how i solve the problem from exercise 6.5 of question no.14, give me the best way to solve the problem of this question 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.

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# 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. Q.14

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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,

AB

^{2}=^{ }AD^{2}+ BD^{2}……………………….(i)AC

^{2}=^{ }AD^{2}+ DC^{2}……………………………..(ii)Subtracting equation

(ii)from equation(i), we getAB

^{2}– AC^{2}= BD^{2}– DC^{2}= 9CD

^{2}– CD^{2}[Since, BD = 3CD]= 8CD

^{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}.