Can someone please prove this?

enter image source here

2 Answers
May 8, 2018

See explanation.

Explanation:

enter image source here
Let angleTCB=x, => angleOCB=90-x,
as OC=OB=r, => angleOBC=angleOCB=90-x,
=> angleCOB=180-(90-x)-(90-x)=2x
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,
=> angleCAB=(angleCOB)/2=x,
given that AB is the diameter, => angleACB=90^@,
=> DeltaACB and DeltaCDB are similar,
=> angleDCB=angleCAB=x

Hence, BC bisects angleTCD

May 8, 2018

See the answer below...

Explanation:

enter image source here

  • 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...

:-)