- #1
tantrik
- 13
- 0
Dear friends,
I am unable to find out the unknown angles for the following triangle which I attached with this post.
View attachment 6092
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.
I am unable to find out the unknown angles for the following triangle which I attached with this post.
View attachment 6092
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.