ABCD is a quadrilateral. P,Q,R, and s are the mid
Points of AB, BC, CD and DA respectively . Show
PR and QS bisects each other.
It is hard for me to tell how rigorous your proof must be.
One of the features of Euclidean geometry is that it is cumulative, that is, you can use the result of one problem as a given in a new problem
To reach this point illustrated by your problem above, you must have at one time proved that :
In any triangle the line joining the midpoints of any two sides is one-half the length and parallel to the third side.
As I said, I don't know whether your method would involve simply using that result, or whether you have to actually prove that as part of the above problem. I would guess you wouldn't have to.
So you can prove that in PQRS , the opposite sides are parallel, thus making it a parallelogram.
Then by drawing diagonals of this parallelogram, you can prove triangles congruent, ASA all over the place, and thus prove the diagonals bisect each other.
(bisect means cutting into two equal parts)
To show that PR and QS bisect each other in quadrilateral ABCD, we will use the concept of midpoints and parallel lines.
First, let's start by drawing a diagram of quadrilateral ABCD.
```
A______B
| |
| |
|______|
D C
```
Now, let's label the midpoints of the sides: P is the midpoint of AB, Q is the midpoint of BC, R is the midpoint of CD, and S is the midpoint of DA.
```
A______B
|P |
| |
|______|Q
D C
S R
```
To show that PR and QS bisect each other, we need to prove two things:
1. PR divides QS into two equal segments.
2. QS divides PR into two equal segments.
Let's start with the first statement: PR divides QS into two equal segments.
To prove this, we will use the concept of parallel lines and corresponding angles.
Since P and Q are midpoints of sides AB and BC, respectively, we can conclude that PQ is parallel to AC. Similarly, since R and S are midpoints of sides CD and DA, respectively, we can conclude that RS is parallel to AC.
Now, by examining triangle AQC, we can observe that angle QAC is congruent to angle CAQ. Similarly, in triangle DRC, angle RDC is congruent to angle CDR.
Since PQ is parallel to AC, angle QAC is congruent to angle PQR (corresponding angles). Similarly, since RS is parallel to AC, angle CDR is congruent to angle CSQ.
Now, considering triangle AQC, we have angle CAQ congruent to angle QAC and angle QAC congruent to angle PQR. Therefore, we can conclude that angle CAQ is congruent to angle PQR.
Similarly, considering triangle DRC, angle CDR is congruent to angle RDC and angle RDC congruent to angle CSQ. Therefore, angle CDR is congruent to angle CSQ.
From the above information, we can see that triangles AQC and DRC are similar by angle-angle similarity.
Since a pair of corresponding angles are congruent, corresponding sides in these triangles are proportional. Corresponding sides are PQ/QS, AC/DR, and AQ/RC.
Since PQ is half the length of AB and QS is half the length of BC, PQ/QS = AB/BC = 1/2. Similarly, AQ/RC = AB/CD = 1/2.
Since the corresponding sides are proportional with a ratio of 1/2, we can conclude that PR divides QS into two equal segments.
Now, let's move on to the second statement: QS divides PR into two equal segments.
To prove this, we will use the same concept of parallel lines and corresponding angles.
Since PQ is parallel to AC, angle PQR is congruent to angle CAQ (corresponding angles). Similarly, since RS is parallel to AC, angle CSQ is congruent to angle CDR.
Now, considering triangle AQC, angle CAQ is congruent to angle PQR and angle CAQ is congruent to angle QAC. Therefore, angle PQR is congruent to angle QAC.
Similarly, considering triangle DRC, angle CDR is congruent to angle CSQ and angle CDR is congruent to angle RDC. Therefore, angle CSQ is congruent to angle RDC.
From the above information, we can see that triangles AQC and DRC are similar by angle-angle similarity.
Since a pair of corresponding angles are congruent, corresponding sides in these triangles are proportional. Corresponding sides are PQ/QS, AC/DR, and AQ/RC.
Since PQ is half the length of AB and QS is half the length of BC, PQ/QS = AB/BC = 1/2. Similarly, AQ/RC = AB/CD = 1/2.
Since the corresponding sides are proportional with a ratio of 1/2, we can conclude that QS divides PR into two equal segments.
Therefore, we have shown that PR and QS bisect each other in quadrilateral ABCD.
After joining AC and using the midpoint of a triangle theorem ,it is easy to prove PQRS is a parallelogram
A property of a parallelogram is that its diagonals bisect each other.
Done.
Write it up according to the method that your instructor wants.