One of the legs of the obtuse isosceles △ABC and △DBE lie on the same line sharing vertex B without overlapping. The sides DE and AC intersect at point F, DC = 12 in, m∠BDF=m∠BCF=30°. Find AE and CE.
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Answer:

  AE = CE = 12 in

Step-by-step explanation:

Angles BDF and BCF are base angles of their respective isosceles triangles. That means the a.pex angle of each, DBF and ABC respectively is ...

  180° -2×30° = 120°

In your diagram, BC is 120° from the -x axis, so is 60° from the +x axis. It bisects the angle DBE. Similarly, BE is 120° from the +x axis, so is 60° from the -x axis. It bisects angle ABC. In an isosceles triangle, the a.pex angle bisector is an altitude and is the perpendicular bisector of the base segment. Any point on that line is equidistant from the base vertices.

Point C lies on the perpendicular bisector of DE, so CD ≅ CE = 12 in.

Point E lies on the perpendicular bisector of AC, so EC ≅ EA = 12 in.

The measurements of interest are ...

  AE = CE = 12 in.

_____

Additional comment

Point F serves no purpose except to confuse the issue. Each of the angles that reference point F could be described equally well using a different point:

  ∠BDF = ∠BDE

  ∠BCF = ∠BCA

This makes it more obvious that the triangles of interest are similar.

Answer:

12

12

Step-by-step explanation:

see the below figure

i got that both triangles are similar

we get that in triangle ABC angle BCF is 30

so angle A is also 30 (isosceles)

so the remaining angle ABC is 120

and angle CBD becomes 180 - 120 = 60

now in triangle BDE angle BDF is 30

so angle BED is also 30

now angle EBC becomes 120 - angle CBD = 120 - 60

and got angles ABE EBC CBD as 60 each

so three 60 degree angles are formed at vertex B as shown in the figure

 tw0 30 30 120 degree triangles are similar

BC becomes perpendicular angular bisector of ED

C is a point on that line which is at equal distance from the base vertices

so CD = CE = 12

here BE is perpendicular bisector of AC

so AE = EC = 12    

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