Question #5050

If a point lies on the bisector of an angle, then the point is equidistant from the sides of the angle.

Expert's answer

Proof:

let the angle be ABC (B is the vertex with the angle). Let its angle bisector be BD. Assume a point E anywhere on the line BD. Join E to BC and BA PERPENDICULARLY because distance is always measured perpendicularly. Now we have two right triangles BAE and BCE.

Now in these two triangles-

Statements Reasons

Angle ABE = Angle CBE & & & & & & & & & & & & & & & & ( angle bisector)

Side BE = Side BE & & & & & & & & & & & & & & & & & & & & & (common line to both the triangles)

Angle BAE = Angle BCE & (right angles)(90*)

Hence by Angle-side-angle CONGRUENCE rule, triangle BAE is CONGRUENT TO triangle BCE.

and by CPCT AE IS EQUAL TO CE. (which is the perpendicular distance).

let the angle be ABC (B is the vertex with the angle). Let its angle bisector be BD. Assume a point E anywhere on the line BD. Join E to BC and BA PERPENDICULARLY because distance is always measured perpendicularly. Now we have two right triangles BAE and BCE.

Now in these two triangles-

Statements Reasons

Angle ABE = Angle CBE & & & & & & & & & & & & & & & & ( angle bisector)

Side BE = Side BE & & & & & & & & & & & & & & & & & & & & & (common line to both the triangles)

Angle BAE = Angle BCE & (right angles)(90*)

Hence by Angle-side-angle CONGRUENCE rule, triangle BAE is CONGRUENT TO triangle BCE.

and by CPCT AE IS EQUAL TO CE. (which is the perpendicular distance).

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