Angles: Very Difficult Problems with Solutions

Solution:
We make line EF longer until it intersects line AB in point M. ∠DEF and ∠AME are corresponding. Since DE and AM are parallel, m∠AME also has 45°. ∠AME and ∠ABC are corresponding. Since MF and BC are parallel, ∠ABC also has 45°. To conclude, two acute angles whose sides are parallel two by two have the same measure.

Solution:
We make side EF longer until it intersects side AB in point M. ∠DEF and ∠AME are corresponding. Since DE and AM are parallel, ∠AME also has 130°. ∠AME and ∠ABC are corresponding. Since MF and BC are parallel, ∠ABC also has 130°. To conclude, two angles whose sides are parallel two by two have the same measure.

Solution:
We make EF longer until it intersects AB at point M. ∠DEF and ∠AMN are corresponding. Since DE and AM are parallel, ∠AMN and ∠DEF are equal. ∠AME and ∠ABC are corresponding. Since MF and BC are parallel, m∠AME = 40°. ∠AMN and ∠AME are adjacent and supplementary. Thus, m∠AMN=140°, so m∠DEF=140°. To conclude, two angles, one of them acute and the other obtuse, whose sides are parallel, are supplementary.

Solution:
Let the obtuse angle at [tex]B[/tex] be [tex]2\varphi[/tex]; the bisector [tex]BC[/tex] divides it into two angles of [tex]\varphi[/tex].
Since [tex]a\parallel b[/tex], the line [tex]BC[/tex] forms with [tex]a[/tex] the same angles as with [tex]b[/tex] (alternate angles): [tex]\varphi[/tex] and [tex]180^\circ-\varphi[/tex].
[tex]\varphi\lt 90^\circ[/tex] (because [tex]2\varphi\lt 180^\circ[/tex]), so [tex]180^\circ-\varphi=125^\circ \Rightarrow \varphi=55^\circ[/tex].
The obtuse angles are [tex]2\times 55^\circ=110^\circ[/tex] and the acute ones are [tex]180^\circ-110^\circ=70^\circ[/tex]: all eight angles are [tex]110^\circ[/tex] or [tex]70^\circ[/tex].

Solution:
Let the transversal meet the lines at [tex]A[/tex] and [tex]B[/tex], let the co-interior angles be [tex]2\alpha[/tex] (at [tex]A[/tex]) and [tex]2\beta[/tex] (at [tex]B[/tex]), and let the bisectors meet at [tex]M[/tex].
In triangle [tex]ABM[/tex]: [tex]\angle MAB=\alpha[/tex], [tex]\angle MBA=\beta[/tex], [tex]\angle AMB=90^\circ[/tex]. The angles of a triangle add up to [tex]180^\circ[/tex]:
[tex]\alpha+\beta+90^\circ=180^\circ \Rightarrow \alpha+\beta=90^\circ \Rightarrow 2\alpha+2\beta=180^\circ[/tex]
The co-interior angles add up to [tex]180^\circ[/tex], so the lines are parallel.