Can someone please prove this?
2 Answers
See explanation.
Explanation:
Let
as
recall that the angle subtended by an arc at the center of a circle is twice the angle subtended by the same arc at the circumference,
given that
Hence,
See the answer below...
Explanation:
- I have connected the center with the tangent, so
bar("OC") is perpendicular tobar("CT") i.e,/_"OCT"=90^@ Delta"ACB" is a semi circled triangle. So,/_"ACB"=90^@
/_ "OCT" = /_"ACB"
=>/_ "OCT"-/_ "OCB"=/_ "ACB"-/_"OCB"
=>/_ "ACO"=/_ "BCT" " "......(1)
Delta"OCD" andDelta "DCB" are right-angled triangles. So,
/_ "COD" + /_ "OCD" = /_ "DCB" +/_ "CBD"=90^@
=>/_ "COD" + /_ "OCD"+/_ "DCB" = /_ "DCB" +/_ "CBD"+/_ "DCB"" " [adding/_ "DCB" both side]
=>/_ "COD" +( /_ "OCD"+/_ "DCB" )=2 /_ "DCB" +/_ "CBD"
=>/_ "COD" + /_ "OCB"= 2/_ "DCB" +/_ "CBD"
/_ "OCB"=/_ "CBD" , becauseDelta"OCB" is a isosceles triangle].
=>/_ "COD" + cancel(/_ "OCB")= 2/_ "DCB" +cancel(/_ "CBD"
=>/_ "COD"=2/_ "DCB"" ".......(2)
- Further,
Delta"ACO" is isosceles triangle, so/_ "OAC"=/_ "ACO" - The sum of two internal angles of a triangle is equal to the opposite external angle.
2/_ "ACO"=/_ "COD"" "....(3) From
(1), (2), (3) we get,
color(red)(ul(bar|color(blue)(/_ "BCD"=/_ "BCT")|) That means,
bar("BC" disects/_ "TCD" Hope it helps...
Thank you...
:-)