3. A line-segment parallel to the bases of any trapezoid, terminating in the sides, and drawn through the intersection of the diagonals, is bisected at that point.

4. In ABC, DE is parallel to the side AC and intersects AB at D and BC at E. If DC and AE intersect at O, prove BO a median of the triangle. Use Ex. 3.

A

B

D

E

0

C

5. If one of the bases of a trapezoid is double the other, the diagonals intersect at a point of trisection of each.

7. If three or more concurrent lines intersect two parallel lines, the

8. If three or more straight lines intersect two parallel lines, making the corresponding intercepted segments of the parallel lines proportional,

the lines are either concurrent or parallel.

HYPOTHESIS. - In a figure lettered as in Ex. 7,

AB BC
EF FG

= etc.

CONCLUSION. — AE, BF, CG, etc., are either concurrent or parallel. SUGGESTION.- Suppose that no two of these lines are parallel and that AE and BF intersect at O. Draw OC, intersecting EF at M. Prove that M and G coincide, and hence that CG passes through 0.

9. The straight line joining the middle points of the bases of any trapezoid and the two non-parallel sides produced are concurrent.

188. Internal and external division of a line-segment. - A linesegment is divided internally into two parts at a point between its extremities, and externally into two parts at a point on

the segment produced through one of its extremities.

Thus AB is divided internally at Minto AM and MB, and is divided externally at N into AN and NB.

189. Theorem. — The bisector of an angle of a triangle divides the opposite side internally into segments proportional to the adjacent sides.

Hypothesis. In △ABC, CD bisects ∠ACB.

Proof. 1. In △ABC, CD bisects ∠ACB.

2. Draw BE || CD and produce AC to meet it at E.

Def. bisector

§ 26

§ 29

Ax. I

§ 74

Ax. XII

190. Theorem. - The bisector of an exterior angle of a triangle that intersects the opposite side divides it externally into segments proportional to the adjacent sides.

Hypothesis. In △ABC, CD bisects the exterior angle at C

and intersects AB produced at D.

Suggestions. Draw BE || CD, meeting AC at E. Prove

AD AC

that

Hence the conclusion may be established if

DB CE

it can be proved that CE = BC.

Write the proof in full.

1. In

EXERCISES

ABC, AB = 5 in., AC = 6 in., and BC = 7 in. Find the

segments into which the bisector of ∠C divides AB.

SUGGESTION. - Let the segment adjacent to AC be x in. Then

2. In the triangle of Ex. 1, find the segments into which the sides AC and BC are divided by the bisectors of the opposite angles.

3. The sides of a triangle are 8 in., 10 in., and 12 in. Find the segments into which the 12-inch side is divided by the bisector of the exterior angle at the opposite vertex.

4. Prove by § 189 that the bisector of the vertical angle of an isosceles triangle bisects the base.

5. When a line-segment is divided internally and externally in the same ratio, it is said to be divided harmonically. Prove that the bisectors of the interior and exterior angles at any vertex of a triangle divide the opposite side harmonically.

6. Prove that a straight line which is drawn through a vertex of a triangle and divides the opposite side internally into segments proportional to the adjacent sides bisects the angle at that vertex.

SUGGESTION. - Using a figure lettered like that of § 189, prove 21 = 22. For this purpose, produce AC to E, making CE = BC. A short indirect proof may be given also.

7. State and prove the converse of the theorem in § 190.

8. The sides of a triangle are a, b, and c. Derive formulæ for the segments into which the side cis divided by the bisector of the opposite angle.

9. Study the adjoining figure, and prove how a given line-segment AB may be divided into segments proportional to two given linesegments m and n, by application of § 189. A (AC = 2m and BC = 2 n.)

n

D

B

191. Projection. - If from the extremities of a given linesegment perpendiculars are drawn to a given line, the segment cut off from the line by the perpendiculars is called the projection of the given line-segment upon the given line.

For example, if ACL MN and BD | MN, the projection of AB upon MN is CD. If A is on MN, and BD .L MN, as in the figure to the right, the projection of AB upon MN is AD. Why?

If the given line-segment AB intersects the line MN, draw a figure to show the projection of AB upon MN.

192. Theorem. - Either leg of a right triangle is a mean proportional between the hypotenuse and its projection upon the hypotenuse.

A

C

D B

Hypothesis. In △ABC, ∠ C is a right angle, and AD and DB are the projections of AC and BC, respectively, upon AB.

angles are first proved similar? It may be proved that AB BC if what triangles are first proved similar?

BC

DB

Write out the complete proof.

193. Corollary. - A chord of a circle is a

C

mean proportional between the diameter drawn from one of its extremities and its projection

upon that diameter.

The proof is left to the student.

EXERCISES

1. In the figure of § 192, if AB = 25 in. and AD = 16 in., find AC and BC.

2. The hypotenuse of a right triangle is 18 in. and a leg is 15 in. Find the projection of the leg upon the hypotenuse.

2

3. In the figure of § 192, prove that AC = AB × AD and BC2 =

AB x DB.