Please prove this also ..........
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Hi pupil here's your answer ::
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Since , OR perpendicular to PQ
➡<ROQ+<ROP = 180° {LINEAR PAIR}
➡90° + <ROP = 180°
➡<ROP = 180°- 90° = 90°
▶<ROS+<POS=90° (ROP = ROS+POS)
▶<POS = 90° - <ROS..........☝
⏩<QOS = <ROS+ <ROQ
⏩<QOS = <ROS + 90°...............✌
subtracting ☝from✌ we get
✔QOS - POS = ROS +90° - (90°-ROS)
✔QOS - POS =ROS +90-90+ROS
✔ QOS - POS = 2ROS
✔
so now
✔
Hence proved
_____________________________
hope that it helps. . . . . .
_____________________________
Since , OR perpendicular to PQ
➡<ROQ+<ROP = 180° {LINEAR PAIR}
➡90° + <ROP = 180°
➡<ROP = 180°- 90° = 90°
▶<ROS+<POS=90° (ROP = ROS+POS)
▶<POS = 90° - <ROS..........☝
⏩<QOS = <ROS+ <ROQ
⏩<QOS = <ROS + 90°...............✌
subtracting ☝from✌ we get
✔QOS - POS = ROS +90° - (90°-ROS)
✔QOS - POS =ROS +90-90+ROS
✔ QOS - POS = 2ROS
✔
so now
✔
Hence proved
_____________________________
hope that it helps. . . . . .
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