∠BED = ∠BDE and E is the midpoint of BC. Prove that AF/CF = AD/BE.

We have

Given: ∠BED = ∠BDE

E is the mid-point of BC.

⇒ BE = EC

Construction: Draw CG ∥ AB.

To Prove:

Proof: Let ∠BED = ∠BDE = x

In ∆BDE,

If ∠BED = ∠BDE, then

BD = BE [∵ The sides opposite to the equal angles in a triangle is always equal]

Note that, in ∆CEG,

∠CEG = x [∵ ∠BED = x; they are vertically opposite angles]

So, note in ∆BED and ∆CEG,

∠BED = ∠CEG

∠BDE = ∠CGE [∵ they are alternate angles]

∴ ∆BED ∼ ∆CEG by AA similarity.

Also, we know BE = EC.

∴ ∆BED ≅ ∆CEG by AAS rule of congruency of triangles.

Hence, ∠CEG = ∠CGE = x

Also, ∠ECG = ∠DBE [∵ they are alternate angles]

Now, for ∆ABC, ∠FCB is an external angle to it.

By property that says that, external angle to a triangle is equal to the sum of interior opposite angles.

⇒ ∠FCB = ∠A + ∠B

Since DB ∥ CG or AB ∥ CG. Then

∠GCF = ∠A [∵ they are alternate angles] …(i)

∠CGF = 180° - ∠CGE

⇒ ∠CGF = 180° - x

In ∆CGF,

∠F = 180° - ∠A – (180° – x)

⇒ ∠F = 180° - ∠A – 180° + x

⇒ ∠F = x - ∠A

Now compare ∆ADF and ∆CGF.

∠F = ∠F [∵ they are same angles]

∠DAF = ∠GCF [from equation (i)]

So, ∆ADF ∼ ∆CGF by AA Similarity rule.

⇒

[∵ CG = CE]

Clearly, .

Thus, proved.