In the given figure, ∠ CEF - ∠ CFE. F is the midpoint of DC. Prove

Q) In the given figure, ∠ CEF = ∠ CFE. F is the midpoint of DC. Prove that $\frac{AB}{BD} = \frac{AE}{ FD}$ .

Ans:

Step 1: From given conditions:

∴ CE = CF (sides of opp. Angles)

2. ∵ F is midpoint of CD

∴ FD = CF

∴ CE = CF = FD ……… (i)

Step 2: Construct to find relation:

Let’s draw DG ǁ BE

∴ by Basic Proportionality Theorem in Δ ABE: $\frac{AB}{BD} = \frac{AE}{GE}$ ………… (ii)

Step 3: establish co-relation:

∴ by Mid Point Theorem in Δ GDC : $\frac{DF}{FC} = \frac{GE}{CE}$

∵ DF = FC ……. from equation (i)

∴ $\frac{DF}{FC}$ = 1 = $\frac{GE}{CE}$

∴ GE = CE

∵ CE = CF = FD ….. from equation (i)

∴ CE = CF = FD = GE ……. (iv)

Step 4: Proving the relation:

∵ $\frac{AB}{BD} = \frac{AE}{GE}$ ……. from equation (ii)

and GE = FD ………… from equation (iv)

∴ $\frac{AB}{BD} = \frac{AE}{FD}$

Hence Proved!

Contact us