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angCAB=angCAP + angPAB
angCAP=1/2CAB
angPAB=1/2CAB
SIMILARLY ..
angDBP=1/2DPA
angPBA=1/2DPA
SO WE CAN SAY.:
angC+angD=180°(INTERIOR ANGLE)
angPAB+angPBA=90(bcouz it is half angC & D)
ang2PAB+ang2PBA=180
in∆apb
ang1=ang2
45°=45°{bcouz we proved earlier}
in∆apb
ang1+ang2+ang3=180
45+45+ang3=180
ang3=90
so we can state..
angC+angD=180
angAPB=90
so ang2APB=angC+angD
angCAP=1/2CAB
angPAB=1/2CAB
SIMILARLY ..
angDBP=1/2DPA
angPBA=1/2DPA
SO WE CAN SAY.:
angC+angD=180°(INTERIOR ANGLE)
angPAB+angPBA=90(bcouz it is half angC & D)
ang2PAB+ang2PBA=180
in∆apb
ang1=ang2
45°=45°{bcouz we proved earlier}
in∆apb
ang1+ang2+ang3=180
45+45+ang3=180
ang3=90
so we can state..
angC+angD=180
angAPB=90
so ang2APB=angC+angD
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