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

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In quadrilateral ABCD, with AB equal to AC, the angles are given as ∠ABD = 60°, ∠ADB = 70°, and ∠BDC = 40°. To find ∠DBC, the properties of isosceles triangles and the sum of angles in a triangle are applied. The calculations reveal that ∠DBC equals 70°. This conclusion is reached by analyzing the relationships between the angles and using the triangle sum theorem. The final answer is ∠DBC = 70°.
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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)
$\therefore \angle DBC=\angle DBE=10^o$
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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