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Isosceles base angles, the midsegment, the medians and the centroid, vertical angles and the exterior angle theorem.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
In this lesson you turn the congruence criteria into results you can use. You prove and apply the base angles theorem for isosceles triangles, the midsegment theorem, the fact that the three medians meet at a centroid that cuts each of them in the ratio two to one, the equality of vertical angles, and the exterior angle theorem. Each one is reached the same way: find two congruent triangles, or two angles supplementary to the same third angle, and then read off the parts you wanted.
You know that the three angles of a triangle add to $180^\circ$ and that angles on a straight line add to $180^\circ$, and you have just met the congruence criteria SSS, SAS, ASA and AAS. That is the whole toolkit: every theorem below is proved either by finding two congruent triangles or by noticing two angles that share a supplement.
Isosceles triangle: one with two sides of equal length. The angle between those two sides is the apex angle; the other two are the base angles.
Median: the segment from a vertex to the midpoint of the opposite side. Every triangle has three.
Centroid: the single point where the three medians meet.
Midsegment: the segment joining the midpoints of two sides.
Vertical angles: the two opposite angles made where two lines cross — they share only the crossing point.
Exterior angle: the angle between one side and the extension of the next. The two interior angles not touching it are its remote interior angles.
Each of these is proved, not measured. Base angles: in an isosceles triangle the two angles opposite the equal sides are equal, because the median from the apex splits it into two triangles congruent by SSS. So an apex of $40^\circ$ leaves $140^\circ$ to share, and each base angle is $70^\circ$. Midsegment: a segment joining the midpoints of two sides is parallel to the third side and half its length, so midpoints on a triangle with third side $16$ give a midsegment of $8$. Medians and the centroid: the three medians meet at one point that cuts each of them in the ratio $2 : 1$ from the vertex, so on a median of $18$ the centroid sits $12$ from the vertex and $6$ from the midpoint. Vertical angles: where two lines cross, opposite angles are equal, because each is supplementary to the same neighbour — if the neighbour is $110^\circ$, both of the others are $70^\circ$. Exterior angle: an exterior angle equals the sum of the two remote interior angles, so interior angles of $50^\circ$ and $60^\circ$ make an exterior angle of $110^\circ$ at the third vertex.
Another way: picture
An isosceles triangle with a segment drawn from the apex to the midpoint of the base. The two halves share that segment, share the base's two equal halves, and have the two equal sides — congruent by SSS, so the base angles match.
Another way: steps
To prove any of these: name the two triangles you think are congruent, list the three parts that match, name the criterion, then read off the corresponding parts you wanted. When no second triangle exists — as with vertical angles — look instead for two angles supplementary to the same third angle.
Stopping before you halve. An apex of $40^\circ$ leaves $140^\circ$, and that $140$ belongs to two angles, so each base angle is $70^\circ$. The leftover is not the answer; it is the answer doubled.
Reading the $2 : 1$ backwards. The centroid is two parts from the vertex and one part from the midpoint, so it is the far end of the median that gets the short piece. A median of $18$ puts the centroid $12$ from the vertex, not $6$.
Making the exterior angle equal to one remote interior angle. It equals their sum. And a midsegment is half the third side, not equal to it — the two mistakes are the same mistake, using one of the parts where the whole was wanted.
The two base angles are equal by the base angles theorem. Call each $b$.
Name the equal pair before writing any equation.
$40 + b + b = 180$, so $2b = 140$.
$b = 70^\circ$.
Check: $70 + 70 + 40 = 180$.
The interior angle at that vertex is $180 - 50 - 60 = 70^\circ$.
Angle sum.
The exterior angle completes a straight line with it: $180 - 70 = 110^\circ$.
And $50 + 60 = 110$ too, which is the theorem in one step.
The centroid cuts it $2 : 1$ from the vertex, so one part is $21 \div 3 = 7$ cm.
Vertex to centroid $= 2 \times 7 = 14$ cm; centroid to midpoint $= 7$ cm.
A triangle has two sides of equal length, and the angle between those two sides measures $58^\circ$. What does each of the other two angles measure?
answer
Two angles of a triangle measure $34^\circ$ and $57^\circ$. At the third vertex, one side is extended beyond the triangle. What is the exterior angle there?
Answer:
In triangle $ABC$ the median from $A$ to the midpoint of $BC$ measures $15$ cm. The three medians meet at the centroid $G$. How long is $AG$?
Answer:
$M$ is the midpoint of side $AB$ of a triangle and $N$ is the midpoint of side $AC$. The segment $MN$ measures $27$ cm. How long is $BC$?
Answer:
A plot of land is fenced on three sides. A gardener wants a straight gravel path running from the middle of one fence to the middle of the fence beside it. She has never measured either of those two fences, but the third fence, the one the path does not touch, is $32$ m long. She owns $18$ m of edging strip. Is that enough to line the path?
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Two straight lines cross. One of the four angles measures $82^\circ$, and so does the angle vertically opposite it. Which reason *proves* that, without anyone measuring the second angle?
You can use the base angles, midsegment, median, vertical angle and exterior angle theorems, and say which congruence criterion each one rests on. Explain why the exterior angle must equal the sum of the two remote interior angles, and where the centroid sits on a median.
8. Your turn: a median of length $21$ cm, and the centroid on it, step 2