An Angle Inside A Regular Pentagon
Answer : Let G G G be reflection of A A A about E F EF EF . Clearly E G = A E = E D EG=AE=ED EG = A E = E D and ∠ G E D = ∠ A E D − ∠ A E F − ∠ F E G = 10 8 ∘ − 2 4 ∘ − 2 4 ∘ = 6 0 ∘ \angle GED = \angle AED - \angle AEF - \angle FEG = 108^\circ - 24^\circ -24^\circ=60^\circ ∠ GE D = ∠ A E D − ∠ A EF − ∠ FEG = 10 8 ∘ − 2 4 ∘ − 2 4 ∘ = 6 0 ∘ . Hence G E D GED GE D is an equilateral triangle. So G D = D E = D C GD=DE=DC G D = D E = D C and ∠ G D C = 4 8 ∘ \angle GDC=48^\circ ∠ G D C = 4 8 ∘ . Hence ∠ D C G = ∠ C G D = 6 6 ∘ \angle DCG=\angle CGD=66^\circ ∠ D CG = ∠ CG D = 6 6 ∘ , ∠ D G E = 6 0 ∘ \angle DGE=60^\circ ∠ D GE = 6 0 ∘ , and ∠ E G F = ∠ F A E = 5 4 ∘ \angle EGF =\angle FAE = 54^\circ ∠ EGF = ∠ F A E = 5 4 ∘ . So angles C G D , D G E , E G F CGD, DGE, EGF CG D , D GE , EGF sum up to 18 0 ∘ 180^\circ 18 0 ∘ . So G G G lies on F C FC FC . So ∠ D C F = 6 6 ∘ \angle DCF = 66^\circ ∠ D CF = 6 6 ∘ and by symmetry ∠ F D C = 6 6 ∘ \angle FDC=66^\circ ∠ F D C = 6 6 ∘ . It foll...