5) So for parallelogram ABCD, ∠B ≅ ∠D, and ∠A ≅ ∠C. Further, ∠B and ∠A are supplementary (i.e., their sum is 180°), and ∠D and ∠C are also supplementary.
So, we have that m∠B = m∠D. Therefore,
[tex]6x-42=3x+15\\3x=57\\x=19[/tex].
Now, let's substitute for x back into the expression for either ∠B or ∠D to find it's angle measure.
m∠B = [tex]6(19)-42=72[/tex]
Now, remember that ∠B or ∠D are supplements of ∠A.
So, m∠B + m∠A = 180°.
That means m∠A = 180° – 72° = 108°.
That seems reasonable, because A appears to be an obtuse angle.
6)
m∠BAE + m∠DAE + m∠ABC = 180°
[tex]3x+6+10x+2+12x-3=180\\25x+5=180\\x=7[/tex]
∠ABC ≅ ∠ADC.
m∠ABC = [tex]12(7)-3=81[/tex]
m∠ADC = 81°.
7)
[tex]AC=5x+1[/tex]
[tex]AE=3x-3[/tex]
[tex]AB=4x+1[/tex]
[tex]AC=2AE\\\\5x+1=2(3x-3)\\\\5x+1=6x-6\\\\7=x[/tex]
[tex]AB=DC[/tex]
[tex]AB=4(7)+1=29[/tex]
DC = 29
