Question

In triangle ABC,Angle B > Angle C. If AM is the bisector of Angle BAC and AN is perpendicular to BC. Prove that Angle MAN=1/2(Angle B - Angle C)

Answer 1

 Here , Let ∠ CAM  =  ∠ BAM  =  x  ( As given AM is angle bisector of ∠ BAC ) , So  ∠ A  = 2 x                  --- ( 1 )

And

∠  ANB  = ∠ ANC  = 90°                 ( As given AN perpendicular of BC )                   --- ( 2 )

Now from angle sum property of triangle we get in ∆ ABC we get 

∠ A +  ∠ B  +  ∠ C  =  180°  , Substitute value from equation 1 we get 

2 x  +  ∠ B  +  ∠ C  =  180° 

2 x  =  180°  - ∠ B  -  ∠ C

x  = 90° - 12 ∠ B  -  1/2∠C                                        --- ( 3 )

Now from angle sum property of triangle we get in ∆ AMC we get 

∠ CAN +  ∠ ANC  +  ∠ C  =  180°  ,

∠ CAM +  ∠ MAN  +  ∠ ANC  +  ∠ C  =  180°  , Substitute value from equation 1 and 2 we get 

x  + ∠ MAN  + 90° + ∠ C  =  180° 

∠ MAN  = 90° - ∠ C -  x   , Now substitute value from equation 3 and get

∠ MAN = 90° - ∠ C - ( 90° - 1/2 ∠ B  -  1/2∠ C )

∠ MAN = 90° - ∠ C -   90° + 1/2 ∠ B  +  1/2∠ C

∠ MAN =1/2 ∠ B  -  12∠ C

∠ MAN =1/2( ∠ B  -  ∠ C )                                                            ( Hence proved )

Answer 2

Answer:

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