wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

In an equilateral triangle ABC, D is point on side BC such that BD = 13 BC. Prove that 9AD2=7AB2

Open in App
Solution

Let E be the midpoint of BC so that BE=EC=BC2
and the length of the sides of the equilateral triangle is x. So, we have

AB=BC=CA=x(1)

BE=EC=x2

And it is given that BD=BC3=x3

also BE=x2

DE=BEBD=x2x3=x6

Now, In the right angle AED

we have AE=3x2

AD2=AE2+DE2=3x24+x236=28x236

36AD2=28x2

Now put x2=AB2 from equation (1), we get

36AD2=28AB2

9AD2=7AB2 [henceproved]

800727_770341_ans_5f81a5450be4482c948a846c35671ee5.png

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Tango With Straight Lines !!
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon