Question

13. The perpendicular from A on side BC of a AABC meets BC at D such that DB = 3CD
Prove that 2ABP = 2ACP + BC?
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Answer 1

Answer

AnswerGiven that in ΔABC,

we haveAD ⊥BC and BD = 3CD

In right angle triangles ADB and ADC,

we haveAB2 = AD2 + BD2 ...(i)

AC2 = AD2 + DC2 ...(ii) [By Pythagoras theorem]

Subtracting equation (ii) from equation (i),

we getAB2 - 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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Answer 2

Answer:

Step-by-step explanation: