SOLUTION: In the figure plane E is parallel to plane F, AB is perpendicular to F and AG is the perpendicular bisector of CD. If m∠C-AB-D = 30, what is m∠BDC? Image: https://gyaz

Algebra ->  Angles -> SOLUTION: In the figure plane E is parallel to plane F, AB is perpendicular to F and AG is the perpendicular bisector of CD. If m∠C-AB-D = 30, what is m∠BDC? Image: https://gyaz      Log On


   

Question 1013107: In the figure plane E is parallel to plane F, AB is perpendicular to F and AG is the perpendicular bisector of CD. If m∠C-AB-D = 30, what is m∠BDC?
Image: https://gyazo.com/d2c76b9ec1433620e10a45ca1461bd86
Last 2 tutors were absolutely useless, they told me I'm wrong, please help me with this, take a look at the gyazo for clarification. Do not ask me what m∠C-AB-D = 30 means. How the hell am I supposed to know? I'm giving a screenshot for a reason. The teacher gave us a packet to finish as a midterm review guide but this question makes no sense.
Answer by MathLover1(20855) About Me  (Show Source):
You can put this solution on YOUR website!

given:
E || F
AB is perpendicular to F, => AB is perpendicular to E because E || F
AG is the perpendicular bisector of CD-> I think it should be BG is the perpendicular bisector of CD

If m < C-AB-D+=+30 (this is angle between line segment AB and plane in which triangle ACD lie), is same as m and AB)
since BG is the perpendicular bisector of CD, means line segment CG=GD, consequently BC=BD; so, triangle BCD is divided into two right triangles by +BG+
then we know that:
1. measure of angles vertex B and G is 90+
2. sides BC and BD are equal in length, then triangles ABG, BCG, and BGD are right triangles which are similar
take a look at similar triangles ABG and BGD:
angles vertex B and G are both 90+
AG corresponds to BD
GD+correspond to AB
+BG+is common side
if AG corresponds to BD and m < ABG+=30 , then the m < GBD+=30 too
consequently
m < BDC+=60