Question

Given: CD is a perpendicular bisector of AB . Prove: Any point on CD is equidistant from the endpoints of AB . Match each statement with its corresponding reason. Drag the items on the left to the correct location on the right.

Answer 1

Answer:

The two triangles are congruent, so any point on CD will be equidistant from endpoints of AB.

Step-by-step explanation:

Let the consider the figure as per the attached image:

AB be a line whose perpendicular bisector line is CD.

CD divides the line AB in two equal line segments making an angle of $90^\circ$ on both the sides as shown in the attached figure.

Let a point on CD be E.

Here, two triangles are formed:

$\triangle AED\ and \ \triangle BED.$

Side ED is common between the two triangles.

Also, Side ED is perpendicular bisector:

$\angle EDA = \angle EDB =90^\circ$

And Sides AD = DB

According to SAS congruence (i.e. Two sides are equal and angle between them is equal):

$\triangle ADE \cong \triangle BDE$

And as per the properties of congruent triangles, all the sides are equal.

$\Rightarrow EA = EB$

EA and EB is the distance of point E on line CD from the endpoints of line AB.

Hence proved that Any point on CD is equidistant from the endpoints of AB .