Question

In an equilateral triangle ABC, D is a point on side BC such that
BD =⅓
BC. Prove that 9AD² = 7AB²​

Answer 1

Step-by-step explanation:

Given-d is the point on side BC, such that BD=1/3Bc

Draw a line AE perpendicular BC

since triangle ABC is an equilateral triangle

therefore AB=BC=CA=a

and BD=1/3BC=1/3a

NOW

CD=BC-BD

=BC-1/3BC

=2/3BC=2/3A

NOW

DE=BC-BD

=1/2a-1/3a

=1/6a

IN TRIANGLE AED (BY PGT)

AD^2=AE^2+DE^2

=AB^2-BE^2+DE^2 (AE^2=AB^2-BE^2)

=a^2-(1/2a)^2+(1/6a)^2

=a^2-1/4a^2+1/36a^2

AD^2=(36-9+1)a^2/36

=28a^2/36

=7/9AB^2

AD^2=7/9AB^2

9AD^2=7AB^2

HENCE PROVED

Answer 2

Answer:

Check

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