Two circles drawn with centres A and B touch each other externally at C, O is a point on the tangent drawn at C, OD and OE are tangents drawn to the circles of centre A and B respectively. If `angle COD=56^(@), angle COE=40^(@), angle ACD=x^(0) and angle BCE =y^(@),` then prove that `OC = OD =OE and x-y=8.`

Explanation

OC and OD are two tangents at the points C and D on the circle with centre at A form and ecternal point O.
Again, OC and OE are two tangents at the point C and E to the circle with centre B from an external point O.
`therefore OC =OE …….(2)` [by theorem -41] ltbr Then from (1) and (2) we get, `OC = OD = OE.`
Now in `Delta A CD, angle ADC=angleACD =x^(@)[because AC= AD]`
`therefore angle ODC =angle OCD =90^(@)-x^(@)[because angle ADO =angle ACO =90^(@)]`
We know that in `Delta OCE, angle OCE + angle COE =180^(@)`
or,` 90^(@)-y^(@)+90^(@) -y^(@)+40^(@)=180^(@) [because angle COE =40^(@)]`
`or, 2y^(@)=40^(@)or, y^(@) =(40^(@))/(2)=20^(@)`
`therefore y=20 .........(4)`
Then `x-y =28 -20 =8` [from (3) and (4) ]
Hence `OC =ODand x-y=8` (Proved)