AXis the bisector of angle BAC,P is any point on AX.Prove that the perpendicular drawn from P toAB& AC are equal.

Answer :

We form our diagram from given information , As :

Here PM perpendicular on line AB and PN perpendicular on line AC . 

So,

PMA  =  PNA  =  90°                                -------------- (  1 )

Now in PMA and PNA  , We get

PAM  =  PAN                                                  ( As given line AX is angle bisector of BCA and P lies on AX )

PMA  =  PNA                                                 ( From equation 1 )

And

PA  =  PA                                                                          ( Common side )

Hence

PMA PNA                                             ( By AAS rule )

So,

PM  =  PN                                                                       (  By CPCT )                                            ( Hence proved )

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