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secondary 3 | A Maths
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Circle
Date Posted: 4 years ago
Views: 211
let X be the end of line DC
&
Y be the end of line DE
angle XCB = angle CAB
angle YEB = angle EAB
since
angle CAB + angle EAB = 180deg
can conclude
angle XCB + angle YEB = 180deg
&
angle DCB = 180deg - angle XCB
angle DEB = 180deg - angle YEB
so
angle DCB + angle DEB
= 180deg - angle XCB +
180deg - angle YEB
= 360deg - (angle XCB + angle YEB)
= 360deg - (180deg)
= 180deg
this means angle DCB & angle DEB are supplementary (i.e. both add up to 180deg)
&
also, this means angle CDE & angle CBE are supplementary .
these conclude that CDEB is a quadrilateral that is inscribed within circle,
i.e. CSEB is cyclic quadrilateral
.
.
.
&
Y be the end of line DE
angle XCB = angle CAB
angle YEB = angle EAB
since
angle CAB + angle EAB = 180deg
can conclude
angle XCB + angle YEB = 180deg
&
angle DCB = 180deg - angle XCB
angle DEB = 180deg - angle YEB
so
angle DCB + angle DEB
= 180deg - angle XCB +
180deg - angle YEB
= 360deg - (angle XCB + angle YEB)
= 360deg - (180deg)
= 180deg
this means angle DCB & angle DEB are supplementary (i.e. both add up to 180deg)
&
also, this means angle CDE & angle CBE are supplementary .
these conclude that CDEB is a quadrilateral that is inscribed within circle,
i.e. CSEB is cyclic quadrilateral
.
.
.
Thank you Snell
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Two properies of circle are used:
❶ Alternate segment theorem
❷ Opposite angles in a cyclic quadrliateral add up to 180°
❶ Alternate segment theorem
❷ Opposite angles in a cyclic quadrliateral add up to 180°
Date Posted:
4 years ago
Thanks
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