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secondary 4 | E Maths
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Please help me with (b). Thank you!
Date Posted: 6 years ago
Views: 247
RQC = PBX = 62°
RQC + PQX = 180°
(angles on a straight line)
Since RQC = PBX,
PBX + PQX also = 180° .
Since PBX and PQX are opposite sides of a quadrilateral, and they add up to 180°, this means that PBXQ is a cyclic quadrilateral.
Hence a circle can be drawn through the 4 points.
RQC + PQX = 180°
(angles on a straight line)
Since RQC = PBX,
PBX + PQX also = 180° .
Since PBX and PQX are opposite sides of a quadrilateral, and they add up to 180°, this means that PBXQ is a cyclic quadrilateral.
Hence a circle can be drawn through the 4 points.
Thank you!
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