Previously, a very difficult sixth-grade competition problem was released: It is known that the two inner angles of the quadrilateral and the adjacent three sides are equal, and the remaining two inner angles are found! Only a very few can do it, and it is almost annihilated! Some people think that the properties of the diamond shape should be used, and it cannot be solved unless it is superstructured! Graph folding + construction of equilateral triangles can be easily solved!
Question 485: As shown in Figure 1,
Figure 1
In the quadrilateral ABCD, AB=BC=CD, ∠B=80°, ∠C=160°, find ∠A.
Analysis: Graph folding + constructing equilateral triangles!
(1) Connecting BD, which can be seen from BC=CD, ∠CBD=∠CDB=10°. See Figure 2
Figure II
(2) Fold △BCD down along the BD, and the folded point C is recorded as C', then ∠CBC'=∠CDC'=20°. As shown in Figure 3
Figure III
③∠ABC'=∠B-∠CBC'=60°。
(4) Connecting AC', it can be seen from BC'=BC=AB and ∠ABC'=60°, △ABC' is an equilateral triangle. Thus ∠BAC' = 60°, AC' = AB. As shown in Figure 4
Figure IV
(5) From AC'=AB=CD=C'D, it can be seen that △AC'D is an isosceles triangle.
(6) Note that ∠AC'D=360°-∠BC'D-∠AC'B=140°, from △AC'D as an isosceles triangle, ∠C'AD=∠C'DA=20°.
(7) Therefore, ∠A=∠BAC'+∠C'AD=60°+20°=80°.
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