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We know that ∠ABD=125º. We also can see that ∠ABD and ∠DBC are supplementary. So, ∠DBC=180º-125º=55º. Now, that we know ∠DBC=55º we can find out that ∠BCD also is 55º because the triangle is isosceles. ∠DBC, ∠BCD, and ∠BDC are in a triangle, so their sum is 180º. So, ∠DBC=180º-55º-55º=70º. Now, that we know ∠DBC=70º we can find out ∠EDF because the two are vertical angles. ∠DBC=∠EDF=70º.
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