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 to#bar("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" # and#Delta "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"# , because#Delta"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...
:-)