Question

Prove that any point on the angle bisector of an angle is equidistant from its arms

Answer 1

Heya User,

--> I had this pic so just uploaded it =_=

--> Consider the angle bisector AD , and D' any point on it ...
--> By arms , I suppose the question means the points E and F, the points of intersection of altitudes from D ...

--> However, as uh can see :-> 
--> In Δs ADE and ADF , AD = AD
          --> ∠EAD = ∠FAD --> { angle bisector =_= }
          --> ∠AED = ∠AFD --> { 90° }

=> ΔAED and ΔAFD are congruent
=> AE = AF  ---> { by c.p.c.t. }

Eh! Now consider any point D' or so I said already ^_^

--> AD' = AD' { common side }
--> AE = AF   { as I proved above ^ ^ }
--> ∠EAD' = ∠FAD' --> Angle bisector

=> ΔAED' and ΔAFD' are congruent for any point D' on AD 
=> ED' = FD' --> { c.p.c.t. } 

Hence, we are done ^_^ 
--> The above result is true for any arbitrary point D' on AD :p

Answer 2

Answer:

ABOVE ANSWER IS CORRET ☝️