IF BCD is an equilateral triangle, BE is angle bisector of angle B, then find FD : CD

Alternate approach – https://www.youtube.com/watch?v=pnn1POhDRjg
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Comments are closed.

  1. DHANANJAY KUMAR 3 years ago

    Thanks sir

  2. Nice sir

  3. ramesh kumar 3 years ago


  4. SHAMIM Khan 3 years ago

    Triangle BCE in semicircle.. So. <c=90.and <DCE=30. And <CDF=120. so <F=30. FD:DC=1:1

  5. SURYA K 3 years ago


  6. SURYA K 3 years ago


  7. Ankit Maurya 3 years ago

    Sir aapki clear voice nhi ati h

  8. Sushil roman 3 years ago

    Love u sir

  9. Mayank Bhati 3 years ago


  10. RJ 14 3 years ago


  11. praveen pal Pal 3 years ago


  12. ANSHUMAN ANAND 3 years ago

    Excellent Sir

  13. Sangam Kumari 3 years ago

    1:1 ans

  14. Maths is my passion 3 years ago

    Sir complicated bana diya aapne

  15. Mohammad Khalid 3 years ago

    Appreciate your methods sir

  16. ASHUTOSH MISHRA 3 years ago

    in an equilateral triagl…Is angle-bisector same as the median..?

  17. Rakesh Gahlawat 3 years ago

    It can also be solved by with out construction .

  18. BADRI GURU 3 years ago

    CD.BC + BE.PE= ?
    P is intersection point of BE and CD
    Please sir

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