Question

Can someone please prove this?

Answer 1

See explanation.

Explanation:

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`

Answer 2

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...

:-)