12 multiple-choice questions, progressively harder.
In triangle ABC, AC is longer than AB. To begin the cutoff proof, D is chosen on ray AC so AD equals AB. Why is D strictly between A and C?
Solution
Correct answer: A
The chosen distance AD equals the positive length AB.
Since AB is less than AC, moving that distance from A along ray AC reaches D before C. No midpoint or angle bisector is assumed.
In the cutoff construction, D lies between A and C and AD equals AB. Which triangle has angle BDA as an exterior angle?
Correct answer: C
For triangle BDC, side DC extends through D along ray DA.
Angle BDA is therefore exterior to triangle BDC, with remote angles DBC and BCD.
In triangle ABC, angle B is larger than angle C. An indirect proof considers the possibility AC<ABAC<ABAC<AB. What contradiction follows?
Correct answer: B
If AB were the longer side, the longer-side result would make its opposite angle C larger than angle B.
That is the reverse of the given strict comparison.
Triangle ABC has angle measures
For which real x do these form a triangle with BC strictly longest?
Correct answer: D
BC is opposite A, so A must exceed B and C.
Comparing A with B gives
Comparing A with C gives
Thus x must exceed 28. All angles are positive exactly when the third angle also remains positive, requiring x less than 50 in this range. Their sum is already 180 degrees.
An isosceles triangle ABC has AB equal to AC and vertex angle
Which value of x gives a valid triangle with base BC strictly longer than either equal side?
The equal base angles each measure
BC is longer than AB when A exceeds C, which has the same measure as B.
Positive base angles also require x less than 80. Of the choices, only 25 satisfies both conditions.
In triangle ABC, angle A is larger than angle B, and angles B and C are equal. Which pair of side comparisons follows?
AC>BCAC>BCAC>BC
AC=ABAC=ABAC=AB
BC>ACBC>ACBC>AC
AB>BCAB>BCAB>BC
AB=BCAB=BCAB=BC
BC=ACBC=ACBC=AC
AC>ABAC>ABAC>AB
The angles opposite BC, AC, and AB are A, B, and C.
A is larger than B, so
B and C are equal, so
H, Q, and R lie in that order on a line. P is outside the line and PH is perpendicular to it. Which strict comparison follows?
Angle PQR is exterior to right triangle PHQ, because QR extends QH through Q.
It exceeds the right angle at H, so angle PQR is obtuse. It is therefore the largest angle of triangle PQR.
Its opposite side PR is longer than PQ:
P is outside a line, H is its perpendicular foot, and Q and R are different points on the line. Suppose
Which interval satisfies exactly these two comparisons?
The first comparison gives
The second gives
Both inequalities are strict, so both endpoints are excluded.
In an existing triangle ABC, AB is 5 units, AC is 8 units, and angle B is 80 degrees. Which statement about angle C must be true?
AC is longer than AB.
Their opposite angles are B and C, so angle B exceeds angle C. The exact measure of C has not been found.
For the converse theorem, suppose angle B is larger than angle C in triangle ABC. Which two alternatives must be ruled out before concluding AC>ABAC>ABAC>AB?
Two real lengths are equal, one is smaller, or one is larger.
To leave AC greater than AB as the only possibility, rule out equality and AC less than AB. Isosceles base angles and the longer-side theorem provide the contradictions.
A triangle has an exterior angle of 70 degrees and a remote interior angle of 25 degrees. How does that exterior angle compare with its adjacent interior angle?
The adjacent interior angle forms a straight angle with the exterior one.
The difference is
The exterior angle is smaller. Its 25-degree remote angle does not change that adjacent-angle calculation.
Triangle ABC has AB>ACAB>ACAB>AC. Which additional fact would force BC to be strictly longest?
The existing comparison makes AB longer than AC.
If angle A exceeds angle C, then their opposite sides satisfy
Combining the comparisons puts BC ahead of both other sides.
Reset this practice set?
This clears every answer you have given and starts the set again from question 1.