The Converse Of The Perpendicular Bisector Theorem In Geometry

Converse of the Perpendicular Bisector Theorem

If a point is equidistant from the endpoints of a segment, then it is on the perpendicular bisector of the segment.If DA = DB, then D is on the perpendicular bisector of AB

The converse of the perpendicular bisector theorem states that if a point lies on the perpendicular bisector of a segment, then it is equidistant to the endpoints of the segment.

In other words, let’s say we have a segment AB, and a point P lies on its perpendicular bisector. If we draw perpendiculars from point P to the endpoints A and B of the segment, then those perpendiculars will be of equal length. This means that point P is equidistant to points A and B, which is what we mean by saying it is equidistant to the endpoints of the segment.

To prove this, we can use the definition of the perpendicular bisector, which states that the perpendicular bisector of a segment divides it into two equal parts. Let’s call the length of the segment AB x. This means that the perpendicular bisector will intersect AB at its midpoint, which we can call point M.

Now, we can draw perpendiculars from point P to points A and B, which we can call PA and PB, respectively. Let’s say the length of PA is a and the length of PB is b.

Since point P lies on the perpendicular bisector, it is equidistant to points A and B. This means that the distance from P to A is the same as the distance from P to B. In other words,

PA = PB

We can now use the Pythagorean Theorem to relate these lengths to the length of the segment AB. Specifically, we can look at the right triangles AMP and BMP (where M is the midpoint of AB).

In triangle AMP, we have:

– AM = x/2 (by the definition of the perpendicular bisector)
– AP = a

In triangle BMP, we have:

– BM = x/2 (by the definition of the perpendicular bisector)
– BP = b

We also know that PA = PB, so we can set these expressions equal to each other:

a = b

Using the Pythagorean Theorem in triangles AMP and BMP, we get:

– AM^2 + AP^2 = PM^2
– BM^2 + BP^2 = PM^2

Since AM = BM and AP = BP, we can rewrite these equations as:

– (x/2)^2 + a^2 = PM^2
– (x/2)^2 + b^2 = PM^2

But we know that a = b, so we can set these expressions equal to each other:

(x/2)^2 + a^2 = (x/2)^2 + b^2

Subtracting (x/2)^2 from both sides, we get:

a^2 = b^2

Since a = b, we can substitute and get:

a^2 = a^2

This is true, so our assumption that point P lies on the perpendicular bisector must be true as well. Therefore, we have proven the converse of the perpendicular bisector theorem.

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