# $ABCD$ is a quadrilateral in which diagonals bisect each other. Show that $B$ & $D$ are equidistant from $AC$

Well , the question is quite simple (I guess so) .But here, I believe that a parallelogram should be a special one i.e , a square or rhombus, whose diagonals bisect at $90°$ because it's clear from the following theorem - The locus of a point equidistant from two fixed points is the perpendicular bisector of the line segment joining two points.

So my question is whether my belief is correct or the question is true for any NORMAL parallelogram also ??

• Every parallelogram has this property. You can prove the headline question with congruent triangles. Note also this is claim about the distances of point B and point D from the diagonal line AC. – Joffan Mar 17 '17 at 14:24
• Don't we have to prove that $AB = BC$ in the headline question ?? (Parallelogram is named clockwise starting from the left most top as A and continues) – Amritanshu Mar 17 '17 at 14:27
• No, that's not required by the question given. – Joffan Mar 17 '17 at 14:29
• Oh thanks !! I got the meaning of the question !! Thanks !! – Amritanshu Mar 17 '17 at 14:30

Since the diagonals bisect each other, $AM=MC$ and $BM=MD$, we have congruent triangles $\triangle ABM \cong \triangle CDM$ and $\triangle ADM \cong \triangle CBM$ and the shape is thus a parallelogram. Now it is clear that the distance of $B$ from $AC$ is the altitude of $B$ in $\triangle MBC$ and the distance of $D$ from $AC$ is the altitude of $D$ in the congruent $\triangle MDA$ so they are equal.