Question

In the above triangle ABC, AB = BC. BD is perpendicular to AC and EC bisects the angle $\angle ACB$. Is the point E equidistant from AB and BC?

Answer 1

Given, BD is perpendicular to AC.

Since AB = BC, $\Delta ABC$ is an isosceles triangle.
$\therefore$ The perpendicular from B on AC will bisect AC.

So BD is the perpendicular bisector of AC.

We know that the locus of a point which is equidistant from two fixed points is the perpendicular bisector of the line segment joining the two fixed points.
Therefore, here, any point on the line BD is equidistant from A and C.
$\therefore$ P is equidistant from A and C.
Now, EC is the angle bisector of $\angle ACB$.

We know that the locus of a point which is equidistant from two intersecting straight lines is a pair of straight lines which bisect the angles between the given lines. Therefore any point lying on the angle bisector of the angle formed by the intersecting lines is equidistant from those two intersecting lines.
Hence every point on EC is equidistant from lines BC and AC.
Therefore E is equidistant from the lines AC and BC.