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 )
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 )