# In the given figure, if AB = AC, then what is the value of ∠ACB – ∠DBF?

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In the given figure, if AB = AC, then what is the value of ∠ACB – ∠DBF?

posted Feb 12

5 deg

Since AB=AC, ACB = ABC= (180-40)/2=70 deg

As pictured, DBF=CBD=(180-50)/2=65 deg

ACB - DBF=70-65=5 deg

consider triangle ABC
given angle BAC=40 degree and AB=AC -------(1)
therefore triangle ABC is an isosceles triangle

We know, in an isosceles triangle, the angles opposite to equal sides are equal.
therefore,
angle ABC=angle ACB -------(2)

But ,
angle ABC+angle ACB+angle BAC=180 degree
2( angle ABC)+ 40 degree=180 degree [by (1) and (2)]
2(angle ABC)=180 degree-40 degree=140 degree
angle ABC=140/2=70 degree

Therefore,
angle ABC=angle ACB=70 degree ----(3)

But clearly we can see that, angle CAB and angle DBF are corresponding angles.
therefore ,
angle ABC= angle DBF=40 [ Reason: corresponding angles are equal ] ----(4)

By (3) and (4),
angle ACB- angle DBF= 70 degree- 40 degree= 30 degree

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