Question 1195065
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Use the diagram posted by the other tutor (copy it to paper so you can follow the discussion below).<br>
Let m(BCA) = x.  Since AD=DG, m(AGD) = x.<br>
Then m(ADG) = 180-2x; so m(BDG) = 2x.<br>
Since DG=GB, m(DBG) = 2x; so m(BGD) = 180-4x.<br>
Then, with m(AGD) = x and m(BGD) = 180-4x, m(BGC) = 3x.<br>
Since GB = BC, m(GCB) = 3x; and that makes m(CBG) = 180-6x.<br>
With m(DBG) = 2x and m(CBG) = 180-6x, m(CBD) = 180-4x; then since BC = CE, m(BEC) = 180-4x; that then makes m(BCE) = 8x-180.<br>
With m(GCB) = 3x and m(BCE) = 8x-180, m(ECG) = 180-5x.<br>
Then CE = EF means m(CFE) = 180-5x; and then m(CEF) =10x-180.<br>
With m(BEC) = 180-4x and m(CEF) = 10x-180, m(FEA) = 180 - 6x.<br>
Finally, EF = FA means m(BCA) = m(FEA): x = 180-6x.<br>
Solving for x....
x = 180-6x
7x = 180
x = 180/7<br>
ANSWER: The measure of angle BAC is x = 180/7 degrees<br>