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如图,已知AB∥DC,AC⊥BC,AC平分∠DAB,∠B=50°,求∠D的大小.

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解:∵AB∥DC(                  ▲                        ),

∴∠B+∠DCB=180°(                  ▲                        ).

∵∠B=(                  ▲                        (已知),

∴∠DCB=180°﹣∠B=180°﹣50°=130°.

∵AC⊥BC(已知),

∴∠ACB=(                   ▲                        )(垂直的定义).

∴∠2=(                  ▲                        ).

∵AB∥DC(已知),

∴∠1=(                   ▲                        )(                   ▲                        ).

∵AC平分∠DAB(已知),

∴∠DAB=2∠1=(                  ▲                        )(角平分线的定义).

∵AB∥DC(已知),

∴(                  ▲                        )+∠DAB=180°(两条直线平行,同旁内角互补).

∴∠D=180°﹣∠DAB=                  ▲                        

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