1. ## Geometric proof

Hello everyone,

Prove: If two straight lines (line segments) bisect each other forming right angles, any point on either of them is equal distance from the extremities of the other (end points) using Euclid’s method and a more modern approach.

The picture would look like the x and y axis, and I need to prove that the distance from point A is equal to point B, right? For example, (-4,0) is the same distance from the origin as (4,0) is.

My attempt:

Would I need these definitions and propositions to complete the proof?

Definition 3=the ends of a line are points

Definition 10=When a straight line standing on a straight line makes the adjacent angles equal to one another, each of the equal angles is right, and the straight line standing on the other is called a perpendicular to that on which it stands
(to prove when bisected it forms a right angle)

Proposition 13=if a straight line stands on a straight line, then it makes either two right angles or angles whose sum equals two right angles (to prove when the lines are bisected they form right angles)

Proposition 7=given two straight lines constructed from the ends of a straight line and meeting in a point, there cannot be constructed from the ends of the same straight line and meeting in a point, there cannot be constructed from the ends of the same straight line, and on the same side of it, two other straight lines meeting in another point and equal to the former two respectively, namely each equal to that from the same end.

2. Lets take line AB as x axis with origin at its mid point.

So general co-ordinates f A abd B can be give as (-k,0) and (k,0)

Now line bisecting AB and perpendicular to it is y axis.
You have to show that any point on y axis is equi distant feom point a and point B.

Any point on y axis can e denoted as (0,p)

so distance of tis point from A is
$\displaystyle \sqrt((0-(-k))^2 + (p-0)^2)$
$\displaystyle = \sqrt(k^2 + p^2)$

Distance of same point from B is
$\displaystyle \sqrt((0-k)^2 + (p-0)^2)$
$\displaystyle = \sqrt(k^2 + p^2)$

Hence this point of y axis is at equal distance from A and B.

3. Hello, Chocolatelover2!

You're misinterpreting the problem . . .

Prove: If two straight line segments bisect each other forming right angles,
any point on either of them is equal distance from the extremities of the other.
Code:
                * C
|
*P
* | *
*   |   *
*     |     *
A * - - - + - - - * B
|O
|
|
|
|
* D

$\displaystyle AB$ and $\displaystyle CD$ are perpendicular bisectors of each other.
. . They intersect at $\displaystyle O.$
Hence: .$\displaystyle \angle COA = \angle COB = 90^o,\;OA = OB,\;OC = OD$

Let $\displaystyle P$ be any point on $\displaystyle CD.$
We want to prove that: $\displaystyle PA = PB.$

In $\displaystyle \Delta POA$ and $\displaystyle \Delta POB\!:\;\;\begin{array}{c}OA \:=\:OB \\ \angle POA \:=\:\angle POB \\ PO \:=\:PO\end{array}$

Hence: .$\displaystyle \Delta POA \cong \Delta POB$

Therefore: .$\displaystyle PA \:=\:PB$

4. Thank you very much

Do you think this would suffice the Euclidean proof?

For, since AB and CD are perpendicular bisectors of each other, they intersect at point E. (definition of bisectors). Thus, ÐPOA=ÐPOB=90°, OA=OB, OC=OD. Let P be any point that is located on CD. Thus, PA=PB (I-6).

I-6=If in a triangle two angles equal one another, then the sides opposite the equal angles also equal one another

Thank you