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Here are a few recommended readings before getting started with this lesson.
In the net, a quadrilateral, the segments divide the sides into eight congruent segments.
The Triangle Midsegment Theorem gives a relationship between a midsegment and a side of a triangle. There too, is an exciting result for quadrilaterals, formed by the midpoints of the sides of a quadrilateral. Illustrated in the diagram are P, Q, R, and S which are midpoints of the sides of the quadrilateral ABCD.
Show that PQRS is a parallelogram, and that PR and QS bisect each other.
Draw a diagonal in quadrilateral ABCD and focus on the two triangles.
Draw diagonal AC of quadrilateral ABCD and focus on the two triangles △ABC and △ADC.
According to the Triangle Midsegment Theorem, both PQ and SR are parallel to the diagonal AC, and they are half the length of AC. That means these midsegments are parallel to each other, and they have the same length.Similarly, PS and QR are also parallel and have the same length.
By definition, when the opposite sides of a quadrilateral are parallel, then it is a parallelogram. Therefore, the quadrilateral PQRS is a parallelogram.
To show that the diagonals PR and QS bisect each other, focus on two of the triangles formed by these diagonals.
These triangles contain the following properties.
Claim | Justification |
---|---|
PQ≅SR | Proved previously |
∠RPQ≅∠PRS | Alternate Interior Angles Theorem |
∠PQS≅∠RSQ | Alternate Interior Angles Theorem |
These claims can be shown in the diagram.
It can be seen that triangles △PQM and △RSM have two pairs of congruent angles, and the included sides are also congruent. According to the Angle-Side-Angle (ASA) Congruence Theorem, the triangles are congruent.The next part of this lesson focuses on triangles. The diagram shows a triangle with one of its angle bisectors drawn. Move the vertices of the triangle and find a relationship between the displayed segment measures.
The relationship stated in the following theorem can be checked on the previous applet for different triangles.
Find the measurement of the segment as indicated in the applet.
In △ABC, segment AD is the angle bisector of the right angle at A, and DE is perpendicular to AC. The length of the legs AB and AC are 5 and 12, respectively.
Find the length of AE. Write the answer in exact form as a fraction.
Start with finding the length of the hypotenuse and the length of CD.
Mark the lengths which were given in the prompt onto the diagram.
The length of the hypotenuse of the triangle can be found using the Pythagorean Theorem.Substitute expressions
LHS⋅12x=RHS⋅12x
Distribute 12
LHS+12x=RHS+12x
LHS/17=RHS/17
Rearrange equation
Substitute expressions
c/da/b=ba⋅cd
Multiply fractions
LHS⋅156(12−z)=RHS⋅156(12−z)
Distribute 65
LHS+65z=RHS+65z
LHS/221=RHS/221
ba=b/13a/13
Rearrange equation
According to the Angle Bisector Theorem, an angle bisector of a triangle divides the opposite side into two segments that are proportional to the other two sides of the triangle. The converse of this statement is also true.
On the diagram, the markers on line AB are equidistant, the circles are centered at A and at B, and C is the point of intersection of the circles.
Show that CD bisects ∠ACB.
Express the lengths of the line segments in terms of the distance between consecutive markers.
The lengths of some line segments can be expressed in terms of the distance between consecutive markers.
Claim | Justification |
---|---|
AD=2 | By counting the markers |
DB=3 | By counting the markers |
AC=4 | Segment AC is a radius of the circle centered at A. Counting markers shows that the radius of this circle is 4 units long. |
CB=6 | Segment BC is a radius of the circle centered at B. Counting markers shows that the radius of this circle is 6 units long. |
These measurements can be indicated on the diagram.
The ratio of two sides of the triangle can be simplified.