Question

If AP bisects angle BAC and M is any point on AP,prove that perpendiculars drawn from M to AB and AC are equal.

Answer 1

given that ,
 ∠2 = ∠3 ( AP is a bisector ) 
const .
    draw M ⊥  on AP and join to AB on P and similarly in other side M ⊥ AP and join that to AC at Q . 
we get , 
  ∠1 = ∠4 = $90^{0}$
To prove -  PM = QM
proof -  in Δ APM and Δ AMQ 
       ∠2 = ∠3 ( GIVEN  )
      AM = AM ( COMMON )
     ∠1 = ∠4 ( by const. )
 by ASA Δ APM congruent to Δ AMQ .
⇒ PM = QM ( C.P.C.T)
 
 
 

Answer 2

Step-by-step explanation:

given that ,

 ∠2 = ∠3 ( AP is a bisector ) 

const .

    draw M ⊥  on AP and join to AB on P and similarly in other side M ⊥ AP and join that to AC at Q . 

we get , 

  ∠1 = ∠4 = 90^{0}900

To prove -  PM = QM

proof -  in Δ APM and Δ AMQ 

       ∠2 = ∠3 ( GIVEN  )

      AM = AM ( COMMON )

     ∠1 = ∠4 ( by const. )

 by ASA Δ APM congruent to Δ AMQ .

⇒ PM = QM ( C.P.C.T)