Question

in the given figure the bisectors of angle <A and angle <B of parallelogram ABCD meet at O. Find the measure of angle<AOB​

Answer 1

Answer:

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Step-by-step explanation:

so here ao is bisector of angle a and bo is bisector of angle b

thus angle dao=angle bao=1/2×angle dab

and angle cbo=angle oba=1/2×angle cba

thus considering angle dao=1/2×angle dab

angle oba=1/2×angle cba (1)

so we know that

adjacent angles of parallelogram are supplementary

thus angle dab+angle cba=180 (2)

so considering triangle aob and applying angle sum property on it

we get

angle oab+angle oba+angle aob=180

ie 1/2×angle dab+1/2×angle cba+angle aob=180

from (1)

ie 1/2×(angle dab+angle cba)+angle aob=180

ie 1/2×180+angle aob=180 from (2)

ie 90+angle aob=180

ie angle aob=90 degrees