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Step-by-step explanation:
BR,AP , DQ , CQ ARE ANGLE BISECTOR
SO THEY WILL BISECT THR RESPECTIVE ANGLE
IN TRAINGLE DQC
D/2 + DQC + C/2 = 180 eq (1)
( sum of angle of traingle is 180)
IN TRAINGLE ARB
A/2 + ARB+ B/2 = 180 eq(2)
ADDING eq1 and eq 2
A/2 +ARB+B/2 + D/2 + DQC + C/2 = 180 + 180
A/2+ B/2 +C/2+D/2+ ARB+DQC =360
A+B+C+D +ARB+DQC = 360 eq 3
2
{ IN QUADRILATERAL ABCD
A + B + C + D = 360} eq 4
putting eq 4 in eq3
360 +ARB +DQC = 360
2
180 +ARB + DQC = 360
ARB + DQC = 180
similarly
P+ S = 180
HENCE PROVED
MAY IT WILL HELP YOU
SORRY FOR ANY MISTAKE
I DON'T HAVE PAPER NOW SO I CAN SOLVE IT IN THIS WAY
THANK YOU
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Step-by-step explanation:
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