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Asked by sdmbotch1123 | 23 Sep, 2023, 06:51: PM
Figure shows two angles ABC and DEF so that their arms ar parallel. AB || DE and BC || EF .
Let BG and EH are bisectors of respective angles.  When arms are parallel, then ABC = DEF .
Also their bisected angles also equal.

Let us extend bisector HE to meet BC at I as shown in figure.

HEF = GBC ( corresponding angles )

Since HEF = GBC , we also get GBC = EIC .
Hence GB || EI or bisectors are parallel .

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Other possibility of two angles with their arms are parallel is shown in above figure.

AB || DE and BC || EF

Bisector of ABC is drawn and is extended so that it meets DE at H and futher extension meets the bisector of DEFat J .
If HBG = θ , then BGH = (180-2θ) because sum of interior angles between parallel line is 180.
Hence we get , BHG = EHJ = θ (  sum of angles of triangle is 180 )
BGH = DEF = 180-2θ .
Hence bisected angles DEI = FEI = 90-θ
Hence EJH = 180 - (90-θ) - θ = 90o
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