Midpoint and tangent–chord proofs
Derive midpoint relationships, distinguish alternate segments and build a non-circular proof with a missing-condition counterexample.
More
Offline pages
Saved on this device only. Connect at least once every 30 days to confirm Premium. Sign-out or a change of account removes saved copies.
Find this activity in your course
On this page
You stopped at:
ContinueRevision Mode is on. Questions and activities are hidden.
A proof begins with given relationships
A diagram helps locate the objects; its appearance does not establish parallel lines or equal lengths. Begin with the givens, identify corresponding parts and state the reason that makes each next relationship valid. A triangle here has three distinct, non-collinear vertices.
In triangle ABC, M and N are the midpoints of AB and AC. Thus M lies between A and B, N between A and C, and AM/AB = AN/AC = 1/2. The included angle at A is shared, so SAS similarity gives triangles AMN and ABC similar, with A ↔ A, M ↔ B and N ↔ C.
Consequently MN/BC = 1/2 and angle AMN equals angle ABC. Those corresponding-angle equalities imply MN is parallel to BC. This derives the midpoint theorem without using parallelism as a premise. The triangles are similar, rather than congruent: their corresponding lengths have scale factor 1/2 instead of 1.
The converse has its own condition. If a line through midpoint M of AB is parallel to BC and meets AC at N, angle matching makes AMN similar to ABC. Hence AN/AC = AM/AB = 1/2, so N is the midpoint of AC. An arbitrary line through M does not supply those angle equalities and need not bisect AC.
A tangent and chord identify a particular angle
If a tangent touches a circle at A and AB is a chord, the angle between a chosen tangent ray and AB equals the angle standing on chord AB in the alternate segment. Name the tangent direction and the circumference vertex; choosing the opposite direction or other segment changes which supplementary angle is involved.
In this configuration, C lies in the alternate segment for angle TAB. If angle TAB is 60°, the tangent–chord theorem gives angle ACB = 60°. A circumference vertex in the other segment needs the supplementary-angle relationship instead of the same equal-angle statement.
The chord is named by its two endpoints A and B. Angle ACB has its vertex at C and arms CA and CB; angle ADB instead has vertex D. Naming the exact three letters prevents the phrase “same chord” from hiding a change of segment.
Check your understanding 1 Check the ray as well as the chord
In the same figure, angle TAB is 60°. Find angle UAB and angle ADB, giving a reason for each. Explain why neither should simply be assigned the value of angle ACB.
Show answer
Angle UAB is 180° − 60° = 120°, because AT and AU are opposite rays on one straight line. Angle ADB is also 120°: it stands on chord AB from the opposite segment to C, so it is supplementary to angle ACB = 60°. The common chord does not make angles in opposite segments equal, and reversing the tangent ray changes its angle with AB.
Keep reasons tied to the exact vertices
For a proof line, name the angles or lengths compared and the condition that permits the theorem. “Angle AMN equals angle ABC, by corresponding angles since MN is parallel to BC” is valid after parallelism is known. The same sentence cannot be the opening justification when parallelism is what must be proved. In the midpoint proof above, similarity supplies the independent angle equality first, then its converse yields parallelism.
“Same angles” alone does not specify a pair or a reason. Similarly, equal-looking sides cannot establish congruence: state an applicable SSS, SAS, ASA/AAS or right-angle–hypotenuse–side criterion with the corresponding parts. SSA alone does not guarantee congruence. Similarity instead requires matching angles or proportional corresponding sides, with the included-angle condition when using SAS.
Check your understanding 2 The scale factor belongs to corresponding sides
M and N are midpoints of AB and AC in triangle ABC. If BC = 14 cm and the area of ABC is 48 cm², find MN, the area of AMN and the area of quadrilateral MBCN, giving reasons.
Show answer
The shared included angle and ratios AM/AB = AN/AC = 1/2 establish SAS similarity. Thus MN = 7 cm. Area scales by (1/2)² = 1/4, so AMN has area 12 cm² and MBCN has area 48 − 12 = 36 cm². Halving area directly would apply a length factor to a two-dimensional quantity.
Do not use the conclusion as a premise
When asked to prove two lines parallel, begin with an independently known equality, similarity or other sufficient condition. A labelled sketch may suggest the result, but the written argument must work for every configuration satisfying its stated givens. Test whether removing a given would break the reasoning rather than treating all nearby facts as interchangeable.
Try it yourself 1
Link the givens into a proof
Problem
Triangle ABC is inscribed in a circle, with AB = AC. AT is tangent at A, with C in the alternate segment for angle TAB. M and N are the midpoints of AB and AC. Prove that AT is parallel to MN, giving reasons for each step. Then decide whether the two midpoint conditions alone would still establish this conclusion if AB = AC were removed; justify your decision.
Write your answer, then compare it with the solution
Hints
Hint 1
Separate the relationships supplied by each group of givens
The midpoint facts relate the small triangle to the whole triangle. The tangent and equal-side facts constrain particular angles at the circumference. Keep the vertex correspondence visible when connecting those relationships.
Show full solution
Establish the direction of MN
Method
Compare triangles AMN and ABC from the midpoint conditions.Reason
The two side ratios are 1/2 and the included angle at A is shared.Working
Answer
SAS similarity gives AMN similar to ABC, hence MN parallel to BC by the corresponding-angle converse. Equivalently, the midpoint theorem can be used directly, with both midpoint conditions stated. Neither argument assumes anything about the tangent.Establish the tangent's direction independently
Method
Connect the isosceles base angles to the named tangent–chord angle.Reason
AB = AC makes angles ABC and ACB equal; the alternate-segment condition identifies angle ACB for chord AB.Working
Answer
The tangent–chord theorem gives angle TAB = angle ACB. The equal sides give angle ACB = angle ABC, so angle TAB = angle ABC. These are equal alternate interior angles for transversal AB, hence AT is parallel to BC.Combine the two established directions
Method
Use the common parallel direction BC.Reason
Each required parallelism has now been justified from the givens.Working
Answer
Since MN is parallel to BC and AT is parallel to BC, AT is parallel to MN. The conclusion follows without reading parallelism from the diagram or presuming it in an angle statement.Test whether the equal-side condition can be omitted
Method
Keep the midpoint conditions while changing the inscribed triangle.Reason
Midpoints establish MN parallel to BC, but do not by themselves constrain the tangent’s direction.Working
Answer
Without AB = AC there is no reason for angles ABC and ACB to be equal. For a counterexample, take an inscribed right-isosceles triangle with AB its diameter and angle ACB = 90°, so angle ABC = 45°. The tangent at A is perpendicular to AB, whereas BC makes 45° with AB. Thus AT is not parallel to BC or to MN, even though M and N are still midpoints. This triangle has AC = BC, rather than AB = AC, so it satisfies exactly the weakened conditions.
Quick check
Compare the correspondence, each angle pair, theorem conditions and the final parallel inference. A different valid proof is welcome when its reasons are explicit. The missing-condition argument must identify a genuine break or counterexample, rather than saying the diagram might look different. These are proof-comparison criteria, without automatic marking.
Keep track of this lesson
Mark the lesson done when you have worked through it.
Lesson progress is optional and kept on this device.
Loading lesson controls…
Your lesson mark records your own study. Checks are kept separately in My Learning.