Q. In the figure, O is the centre of the circle & angle BCO= 40°.Find x, y and z. DO and CE are both perpendicular to AE.

NOTE: PLEASE DON'T PROVIDE LINKS..

Answer :

We have our diagram , As :

Here we have joined AC and OB  ,

In $∆$ OBC

OC  =  OB                                      ( Radius of circle )

$\angle$OBC  = $\angle$ OCB = 40$°$    ( From base angle theorem , As OC  =  OB )

So, from angle sum property we get

$\angle$ OBC + $\angle$ OCB + $\angle$ BOC = 180$°$  , Substitute values , we get

40$°$ + 40$°$ +  $\angle$ BOC = 180$°$

 $\angle$ BOC = 100$°$
And
we know from theorem " The central angle subtended by two points on a circle is twice the inscribed angle subtended by those points.  "
So,
$\angle$ BOC =  2 $\angle$ BAC , So

2$\angle$ BAC =  100$°$

$\angle$ BAC =  50$°$
And

As we know $\angle$ AEB  =  90$°$ , So OE perpendicular on BC , and we know a line from center to any chord is perpendicular then that line also bisect chord )
SO,
CE  =  BE                            ---- ( 1 )

In $∆$ ABE and $∆$  ACE

CE  =  BE                              (  From equation )

$\angle$ AEB  =  $\angle$ AEC  =  90$°$     ( Given )
And
AE  =  AE                          ( Common side )

Hence $∆$ ABE $\cong$ $∆$  ACE     ( By RHL rule )
So,
$\angle$ BAE  =  $\angle$ CAE                    ( From CPCT )

And
$\angle$ BAC  =  $\angle$ BAE  + $\angle$ CAE  , Substitute values , we get

x  + x  =  50$°$

2x  = 50$°$

x  =  25$°$                                         ( Ans )
From angle sum properties of triangle , we get in $∆$ ABE
$\angle$ AED  +  $\angle$ BAE  +  $\angle$ ABE  =  180$°$  , Substitute all values , we get

90$°$  +  25$°$  + $\angle$ ABE  =  180$°$ 

$\angle$ ABE  =  65$°$ 
And
$\angle$ COD  =  $\angle$ OCB      = 40$°$                          ( Alternate interior angles As given OD  | | BC and take OC as transversal line )

And

we know from theorem " The central angle subtended by two points on a circle is twice the inscribed angle subtended by those points.  "
So,
$\angle$ COD =  2 $\angle$ CBD , So

2y  = 40$°$    

y  = 20$°$                                                                ( Ans )

So,
$\angle$ ABD  =  $\angle$ ABE  - y  =   65$°$  -  20$°$  =  45$°$

And
Quadrilateral ABDF is cyclic quadrilateral , And we know in cyclic quadrilateral opposite angles are supplementary , So

$\angle$ ABD  +  $\angle$ DFA  =  180$°$ , Substitute values , we get


45$°$  + z  = 180$°$ 

So,

z =  135$°$                       ( Ans )