If ABC and PQR are two parallel line, then prove that the bisector of ∠ABQ, ∠CBQ, ∠RQB, and ∠PQB from a rectangle.
Answers
Proved that the bisector of ∠ABQ, ∠CBQ, ∠RQB, and ∠PQB from a rectangle.
Step-by-step explanation:
ABC and PQR are two parallel line,
∠ABQ + ∠CBQ = 180° ( straight line)
∠PQB + ∠RQB = 180° ( straight line)
∠PQB + ∠ABQ = 180° ( as ABC ║ PQR)
∠ACQ + ∠RQB = 180° ( as ABC ║ PQR)
Now let say bisector of ∠ABQ, ∠CBQ, ∠RQB, and ∠PQB
meet at MN to form Quadrilateral BMQN
∠MBN = ∠MBQ + ∠NBQ
=> ∠MBN = (1/2)∠ABQ + (1/2)∠CBQ
=> ∠MBN =(1/2)(∠ABQ + ∠CBQ)
=> ∠MBN =(1/2)( 180°)
=> ∠MBN = 90°
Similalry ∠MQN = ∠MQB + ∠NQB = (1/2) (∠PQB + ∠RQB) = 90°
Let draw a line MS Parallel to ABC & PQR
then ∠QMB = ∠QMT + ∠BMT
∠QMT = ∠PQM ( as PQ ║ MT)
∠BMT = ∠ABM ( as AB ║ MT)
∠QMB = ∠PQM + ∠ABM
=> ∠QMB = (1/2)∠PQB + (1/2) ∠ABQ
=> ∠QMB = (1/2) (∠PQB + ∠ABQ )
=> ∠QMB = (1/2)( 180°)
=> ∠QMB = 90°
Similarly ∠QNB = 90°
as all four angles = 90°
Hence BMQN is a rectangle
Hence proved that the bisector of ∠ABQ, ∠CBQ, ∠RQB, and ∠PQB from a rectangle.
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