Question

ABC is an equilateral triangle, P is a point in bc such that BC:PC=2:1 PROVE that 9AP2=7AB2
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Answer 1

Answer:

ANSWER

From ΔABM,

Let AM⊥BC

AB

2

=BM

2

+AM

2

=(

2

3x

)

2

+AM

2

=

4

9x

2

+(AP

2

−MP

2

) ...[from ΔAMP]

=

4

9x

2

+AP

2

−(

2

x

)

2

AB

2

=AP

2

+2x

2

AB

2

−2x

2

=AP

2

9AB

2

−18x

2

=9AP

2

....[multiplying by 9]

9AB

2

−2(9x

2

)=9AP

2

9AB

2

−2(3x)

2

=9AP

2

9AB

2

−2(AB)

2

=9AP

2

7AB

2

=9AP

2

Step-by-step explanation:

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Answer 2

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From ΔABM,

Let  AM⊥BC

AB2=BM2+AM2

=(23x)2+AM2

=49x2+(AP2−MP2)     ...[from  ΔAMP]     

=49x2+AP2−(2x)2

AB2=AP2+2x2

AB2−2x2=AP2

9AB2−18x2=9AP2      ....[multiplying  by  9]

9AB2−2(9x2)=9AP2

9AB2−2(3x)2=9AP2

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