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29 votes
29 votes

In the figure below, $\angle DEC + \angle BFC$ is equal to _____

  1. $\angle BCD - \angle BAD$
  2. $\angle BAD + \angle BCF$
  3. $\angle BAD + \angle BCD$
  4. $\angle CBA + \angle ADC$
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4 Answers

Best answer
39 votes
39 votes


$\angle z = \angle x + \angle y$

This is a triangle property that tells us that  "the exterior angle of a triangle is the sum of the two opposite interior angle."

Similarly,

$\angle y = \angle a + \angle b$

By substituting, we get, 

$\angle Z = \angle x + \angle a + \angle b$.

Now $\angle Z = \angle BCD$ 
$ \angle b = \angle BAD \\ \angle BCD - \angle BAD =\angle Z - \angle b \\ \qquad \qquad =\angle x + \angle a + \angle b  - \angle b \\ \qquad \qquad =\angle x + \angle a \\ \qquad \qquad = \angle DEC + \angle BFC $

So, option A is the answer.

edited by
71 votes
71 votes

$\angle E + \angle F = ? \\ \alpha + q + E = 180\rightarrow(1) \\ \alpha + \beta + F = 180\rightarrow(2) \\ \alpha + \beta + p + q = 360\rightarrow(3) \\ \text{Equation} \: (1) + (2) = (3) \\ \alpha + q + E + \alpha + \beta + F = \alpha + \beta + p + q \\  \therefore E + F = p - \alpha$

Thus, Correct answer  is option $A.$

edited by
4 votes
4 votes
∠ DEC + ∠ BFC = {180 - (∠BAD+∠ABC)} + {180 - (∠BAD+∠ADC)}

(expanded as all the angles in options are of inner quadrilateral, except ∠BCF; Now lets try leaving ∠BAD and reducing all other terms)

= {360 - (∠BAD+∠ABC+∠ADC)} - ∠BAD

= ∠BCD - ∠BAD
1 votes
1 votes


→ ∠1 = ∠5 + ∠4 ………(i)
According to triangular property:
Angle of exterior = sum of interior angles
→ ∠4 = ∠3 + ∠2 ……….(ii)
By substituting (ii) in (i)
→ ∠1 = ∠5 + ∠3 + ∠2
→ ∠1 = ∠BCD
∠2 = ∠BAD
→ ∠BCD - ∠BAD = ∠1 - ∠2
= ∠5 + ∠3 + ∠2 - ∠2
= ∠ 5 + ∠3
= ∠DEC + ∠BFC

Answer:

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