In Δ ABC, D,E and F are taken on AB, AC and BC respectively, so that EF=CF and DF=BF. If ∠ A=40 ∘ , then find ∠ DFE? Options: A . &nbsp100∘ B . &nbsp120∘ C . &nbsp90∘ D . &nbsp140∘

Answer: Option A : A Option (a) Take DBF = a ∘ and ECF = b ∘ . 40 + a ∘ + b ∘ = 180 ⇒ a ∘ + b ∘ = 140 DFB + EFC + DFE = 180 ⇒ 180 – 2a ∘ + 180 – 2b ∘ + DFE = 180 ⇒ 180 ∘ – 2(a + b) = - DFE DFE = 280 – 180 = 100 ∘

In Δ ABC, D,E and F are taken on AB, AC and BC respectively, so that EF=CF and

Question

In $Δ$ ABC, D,E and F are taken on AB, AC and BC respectively, so that EF=CF and DF=BF. If $∠$ A=40 $^{\circ}$ , then find $∠$ DFE?

Options:

A . 100∘

B . 120∘

C . 90∘

D . 140∘

Answer: Option A
:
A

Option (a)
Take DBF = a

^{\circ}

and ECF = b

^{\circ}

.
40 + a

^{\circ}

+ b

^{\circ}

= 180

\Rightarrow

^{\circ}

+ b

^{\circ}

= 140
DFB + EFC + DFE = 180

\Rightarrow

180 – 2a

^{\circ}

+ 180 – 2b

^{\circ}

+ DFE = 180

\Rightarrow

180

^{\circ}

– 2(a + b) = - DFE
DFE = 280 – 180 = 100

^{\circ}

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