Let CAB be a 20-80-80 triangle, what is the value of DEA? (EAB = 70, DBA = 60)
I tried creating a parallel of BC passing through D but I didn't get anything. I also tried creating an equilateral triangle with AD as its side but it also did not work.
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3This is essentially a duplicate of this question and/or this one, but since those don't have accepted answers, closing as duplicate isn't allowed. There are likely other copies of this question floating around on MSE. (Note that there's also a version with a $50^\circ$ angle.) This is a variants of Langley's Adventitious Angles problem. – Blue Sep 12 '20 at 06:38
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- Calculate some known angles:
$$ACB = 180-(10+70)-(60+20) = 20°$$
$$AEB = 180-70-(60+20) = 30°$$
- Draw a line from point D parallel to AB, labeling the intersection with BC as a new point F and conclude:
$$\triangle DCF \cong \triangle ACB$$
$$CFD = CBA = 60+20 = 80°$$
$$DFB = 180-80 = 100°$$
$$CDF = CAB = 70+10 = 80°$$
$$ADF = 180-80 = 100°$$
$$BDF = 180-100-20 = 60°$$
- Draw a line FA labeling the intersection with DB as a new point G and conclude:
$$\triangle ADF \cong \triangle BFD$$
$$AFD = BDF = 60°$$
$$DGF = 180-60-60 = 60° = AGB$$
$$GAB = 180-60-60 = 60°$$
$\triangle DFG$ (with all angles 60°) is equilateral
$\triangle AGB$ (with all angles 60°) is equilateral
$\triangle CFA$ with two 20° angles is isosceles, so $FC = FA$
Draw a line CG, which bisects ACB and conclude:
$$\triangle ACG \cong \triangle CAE$$
$$FC-CE = FA-AG = FE = FG$$
$$FG = FD, so FE = FD$$
- With two equal sides, DFE is isosceles and conclude:
$$DEF = 30+x = (180-80)/2 = 50$$
Soumyadwip Chanda
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≡ represents "is congruent to"... So wherever you see ≡ , I am talking about triangles and at other places, it is about angles or sides. A MathJax edit would be appreciated :) – Soumyadwip Chanda Sep 12 '20 at 06:44
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What symbol would you like to use for "is congruent to"? Would you like $\equiv$? You get that by using \equiv. – Gerry Myerson Sep 12 '20 at 07:04
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