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I n   F i g .   9 . 42 ,   s i d e   B C   o f   A B C   i s   p r o d u c e d   t o   p o i n t   D   s u c h   t h a t   b i s e c t s   a t   A B C   a n d   A C D   m e e t   a t   p o i n t   E .   I f   B A C = 68 ° ,   f i n d   B E C .

Dear Student ,
Please find below the solution to the asked query :

BAC=68°In ABC,BAC+ABC+BCA=180°       Angle Sum Property 68° + ABC+BCA=180°ABC+BCA=112°   ... iBCA+ACD=180°                     Linear Pair ACD=180°-BCADividing both sides by 2, we getACD2 = 180°-BCA2ACE=180°-BCA2      ... iiIn BCE,EBC + BCE  + BEC=180°         Angle Sum Property ABC2 + BCA + ACE + BEC=180°          Since EBC=ABC2 , As BE is angle bisector ABC2 + BCA + 180°-BCA2 + BEC=180°            Using ii ABC+2 BCA +180°-BCA 2 + BEC=180° ABC+ BCA +180°2 + BEC=180°112°+180°2 + BEC=180°               Using i 146° + BEC=180°  BEC=34°       ANS...
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