Find unknown angles of a triangle!!!

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    Dear friends,

    I am unable to find out the unknown angles for the following triangle which I attached with this post.

    Name:  Math pic.jpg
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    Angle BAD and angle BCD are the unknown angles need to be calculated. Given that lines AB=BC=CD and angle CDE = 108 degrees

    From my calculations: angle ADC = 180 - 108 = 72 degrees (angles on a straight line)

    angle BAD + angle BCD = 108 degrees (exterior angle of a triangle = sum of interior opposite angles)

    I could not proceed any further beyond this. I thought line DB (median) is bisecting angle ADC since line DB is bisecting line AC (AB=BC) but this only happens in case of isosceles and equilateral triangles.

    I am very much stuck here and seek your kind suggestions here. I am also suspecting something could be wrong in the diagram of the triangle or may be the unknown angles. Let me know where I am wrong. Thanks in advance.
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    you've stated yourself that ' angle BAD + angle BCD = 108 degrees '

    angle BCD is a right-angle (90 degrees), from this you can work out angle BAD
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    (Original post by tantrik)
    Dear friends,

    I am unable to find out the unknown angles for the following triangle which I attached with this post.

    Name:  Math pic.jpg
Views: 22
Size:  8.3 KB

    Angle BAD and angle BCD are the unknown angles need to be calculated. Given that lines AB=BC=CD and angle CDE = 108 degrees

    From my calculations: angle ADC = 180 - 108 = 72 degrees (angles on a straight line)

    angle BAD + angle BCD = 108 degrees (exterior angle of a triangle = sum of interior opposite angles)

    I could not proceed any further beyond this. I thought line DB (median) is bisecting angle ADC since line DB is bisecting line AC (AB=BC) but this only happens in case of isosceles and equilateral triangles.

    I am very much stuck here and seek your kind suggestions here. I am also suspecting something could be wrong in the diagram of the triangle or may be the unknown angles. Let me know where I am wrong. Thanks in advance.
    what kind of triangle is BCD and what are the angles in side it.
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    (Original post by gdunne42)
    what kind of triangle is BCD and what are the angles in side it.
    The math teacher did not provide any other information related with the diagram which makes it difficult to solve using geometry principles.
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    EDC = 108, That means CDA=72.
    Given that CDA = 72 and DCA=90, you can find CAD = 180-90-72 = 18.


    The small perpendicular lines on the line segments, AB, BC and DC, means they are all the same length which means, DBC = BDC = 45.

    ABD = 180-45 = 135
    BDA = 72-45 = 27
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    (Original post by tantrik)
    The math teacher did not provide any other information related with the diagram which makes it difficult to solve using geometry principles.
    Assuming BCD is a right angle, You are told two sides of that triangle are equal in length, it is a right angled isosceles triangle so the other two angles in it are 45 degrees.
    Ii it's not a right angle then you have a problem

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