Question

In a ∆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

Answer:

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  -  12∠ 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° - 12 ∠ B  -  12∠ C )

∠ MAN = 90° - ∠ C -   90° + 12 ∠ B  +  12∠ C

∠ MAN =12 ∠ B  -  12∠ C

∠ MAN =12( ∠ B  -  ∠ C )                                                            ( Hence proved)

Tnx for reading

Mark as BEST...

Answer 2

Answer:

hope this helps you :)  :)

Step-by-step explanation: