In a quadrilateral ABCD, AB=AD and BC=DC; and DP is the smallest distance drawn from D on AC. Let’s prove that B, P and D are collinear.

Let us draw a figure for the question given.

ABCD is the required quadrilateral.

It is given that,

AB = AD and BC = CD.

AC is the diagonal of the quadrilateral.

DP is the shortest distance drawn from point D to AC.

Since it is given that AB = AD which means the point A is equidistant from point B and D that is point A is lying at equal distances from point B and D which means that A is a point on the right bisector of segment BD where BD is another diagonal of the quadrilateral.

Similar is the condition for C where C is equidistant from B and D and hence C is lying on the right bisector of segment BD.

This means that AC and BD intersect to give right angles [as both AC and BD are diagonal of a quadrilateral].

So it means that P which is the point on AC drawn from D is the intersection point of the two diagonals.

So it can observed that point P is lying on the diagonal BD.

Hence point B, P and D are collinear that is they are lying on the same line.