12 multiple-choice questions, progressively harder.
In nondegenerate triangle ABCABCABC, DDD lies between A,BA,BA,B, EEE between A,CA,CA,C, and DE∥BCDE\parallel BCDE∥BC. Which part-to-part equality follows?
Solution
Correct answer: B
The side-splitter theorem compares the first part with the leftover part on both sides. Thus AD/DB equals AE/EC.
If AD:DB=2:9AD:DB=2:9AD:DB=2:9, what fraction of the whole side ABABAB is ADADAD?
Correct answer: D
The whole side contains both ratio parts.
ADAB=22+9,=211.\begin{aligned}\frac{AD}{AB}&=\frac{2}{2+9},\\&=\frac 2{11}.\end{aligned}ABAD=2+92,=112.
For the ordinary side-splitter converse in a nondegenerate triangle ABC with AD/DB=AE/ECAD/DB=AE/ECAD/DB=AE/EC, which point-location condition is required?
Correct answer: A
The converse in this lesson concerns a nondegenerate triangle with D between A and B and E between A and C. The positive interior division ratios then determine a unique parallel.
A segment joins the midpoints of two sides of a triangle. Which pair of conclusions is guaranteed?
Correct answer: C
The midsegment theorem supplies both direction and length. It is parallel to the third side and has half its length.
A triangle has third side length 383838 cm. What is the length of the midsegment parallel to it?
A midsegment is half its parallel side.
D,E,F are the midpoints of AB,AC,BCAB,AC,BCAB,AC,BC respectively. Which ordered statement always gives the natural central-to-outer matching?
DE matches CB, EF matches BA, and FD matches AC. Their shared endpoints give D to C, E to B, and F to A.
Three parallel cross-lines meet transversals in corresponding order A,B,CA,B,CA,B,C and P,Q,RP,Q,RP,Q,R. Which equality follows?
AB and PQ are the gaps between the first two cross-lines. BC and QR are the gaps between the last two. Compare them in the same order.
Why must the compass opening stay fixed when marking equal auxiliary steps?
Each step is copied with the same compass opening, so all auxiliary step lengths are equal. The opening can be any convenient positive length.
A construction marks nine equal auxiliary steps and joins the last point to B. The parallel through step 4 meets AB at X. What is AX:XB?
Four steps belong to the first part and the remaining five to the leftover part.
AX:XB=4:(9−4),=4:5.\begin{aligned}AX:XB&=4:(9-4),\\&=4:5.\end{aligned}AX:XB=4:(9−4),=4:5.
Which tools are enough for the exact segment-division construction taught here?
The compass copies equal or supplied lengths; angle copying constructs parallels. The unmarked straightedge draws the required lines without measuring a quotient.
A parallel to BC through the midpoint D of AB meets AC at E in a nondegenerate triangle. What follows?
The division AD:DB is 1:1. The direct side-splitter theorem gives AE:EC=1:1AE:EC=1:1AE:EC=1:1, so E is the midpoint of AC.
To construct a ratio from two supplied positive segment lengths a and b, what is copied consecutively on the auxiliary ray?
Copy the supplied segments in order, join the last point V to B, and draw the parallel through U. Side-splitter transfers AU:UV to AX:XB.
Reset this practice set?
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