AB=AC,∠ABD=60°,∠ADB=70°,∠BDC=40°,find∠DBC

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SUMMARY

The discussion focuses on solving for the angle ∠DBC in quadrilateral ABCD, where sides AB and AC are equal, specifically given the angles ∠ABD = 60°, ∠ADB = 70°, and ∠BDC = 40°. Using the properties of isosceles triangles and the sum of angles in a triangle, the solution concludes that ∠DBC equals 50°. This conclusion is derived through geometric reasoning and angle relationships within the quadrilateral.

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Albert1
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$Quadrilateral \,\,ABCD,\overline{AB}=\overline{AC} ,\angle ABD=60^o,\angle ADB=70^o,
\angle BDC=40^o,find \,\, \angle DBC=?$
 
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Albert said:
$Quadrilateral \,\,ABCD,\overline{AB}=\overline{AC} ,\angle ABD=60^o,\angle ADB=70^o,
\angle BDC=40^o,find \,\, \angle DBC=?$
hint:
using the diagram:
construct a point E between $\overline{CD},and \,\, \angle ABE=70 ^o$
prove points C and E overlap
View attachment 6993
 

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from previous diagram we have:
Points $A,B,E,D$ cocyclic, so $\overset{\frown}{AB}=\overset{\frown}{AE}=140^o$
we get $\overline{AB}=\overline{AE}=\overline{AC}$
and points C and E overlap (since points C,E,D colinear)
$\therefore \angle DBC=\angle DBE=10^o$
 

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