angle bed=angle bde and e is the mid point of bc prove that af/cf=ad/be
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Answered by
338
See diagram...
Let ∠BDE = ∠BED = x°.
Hence, in ΔBDE, we have BD = BE = BC/2.
Now Draw CG parallel to AB.
∠CEG = x° (vertical angle).
ΔBED and Δ CEG are similar as CE || BE, CG || BD, EG || ED.
Since BE = EC, They are congruent.
Hence, ∠CEG = ∠CGE = x° and ∠ECG = ∠B.
Now for the ΔABC, ∠FCB is external angle => ∠FCB = ∠A+∠B
Hence, ∠GCF = ∠A.
Further, ∠CGF = 180° - ∠CGE = 180° - x°.
In ΔCGF , ∠F = 180° - ∠A - (180° - x°) = x° - ∠A
Now compare ΔADF and Δ CGF: Since all corresponding angles are same they are similar triangles.
=> AF/AD = CF/CG = CF / CE = CF / BE
=> AF / CF = AD / BE
Let ∠BDE = ∠BED = x°.
Hence, in ΔBDE, we have BD = BE = BC/2.
Now Draw CG parallel to AB.
∠CEG = x° (vertical angle).
ΔBED and Δ CEG are similar as CE || BE, CG || BD, EG || ED.
Since BE = EC, They are congruent.
Hence, ∠CEG = ∠CGE = x° and ∠ECG = ∠B.
Now for the ΔABC, ∠FCB is external angle => ∠FCB = ∠A+∠B
Hence, ∠GCF = ∠A.
Further, ∠CGF = 180° - ∠CGE = 180° - x°.
In ΔCGF , ∠F = 180° - ∠A - (180° - x°) = x° - ∠A
Now compare ΔADF and Δ CGF: Since all corresponding angles are same they are similar triangles.
=> AF/AD = CF/CG = CF / CE = CF / BE
=> AF / CF = AD / BE
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kvnmurty:
clik on thanks... select best ans..
Answered by
62
Here is you answer please refer the diagram too.
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