The fourth congruence theorem and the converse of Pythagoras · Grade 10 · 5 units
In The fourth congruence theorem and the converse of Pythagoras (Grade 10 · 5 units) you practise 4 skills: The fourth congruence theorem · Two triangles from the same facts · The converse of Pythagoras’ theorem · The circle round a right triangle. The fourth congruence theorem: if two sides of one triangle are equal to two sides of another, and the angles opposite the longer of the two sides are equal, the triangles are congruent. When the equal angles face the shorter side, you cannot conclude that the triangles are congruent. There may be two different triangles, and then the angles facing the longer sides add up to 180°. A simple example: a line from the apex of an isosceles triangle to a point on the base that is not its midpoint splits it into two such triangles, which are not congruent.
The converse of Pythagoras’ theorem: a triangle in which the squares of two sides add up to the square of the third side has a right angle, and the right angle faces the third side. In a right triangle the median to the hypotenuse is half the hypotenuse, so the midpoint of the hypotenuse is the centre of the circumscribed circle. The converse holds too: if the median to a side is half that side, the angle facing it is a right angle. A model proof of the fourth congruence theorem: lay the second triangle on the first so that the equal shorter sides coincide, and so do the equal angles at their end. The third side of the second triangle then runs along the same ray, and the end of its longer side lands on that ray, as far from the shared vertex as the longer side of the first triangle. If it does not land on the first triangle’s vertex, an isosceles triangle appears whose base angles are acute. The angle next to one of them is obtuse, and an obtuse angle faces the longest side of a triangle. That would make the shorter side longer than the longer side, which is impossible. So the vertices coincide, and the triangles are congruent.
Tip: Mark the longer of the two equal sides, and check whether the equal angle faces it.
What you learn
The fourth congruence theorem
For example: In triangles ΔABC, ΔDEF we know that AB = DE = 9, AC = DF = 7, ∢C = ∢F . Are the triangles congruent, and by which theorem?
- Cannot tell
- By the fourth congruence theorem
- By ASA
- By SAS
Answer and solution
The answer: By the fourth congruence theorem
The angle ∢C faces side AB , the longer of the two given sides: 9 > 7 . So the triangles are congruent by the fourth congruence theorem.
Two triangles from the same facts
For example: In triangles ΔABC, ΔDEF we know that AB = DE, AC = DF, AB > AC, ∢B = ∢E = 20° . The triangles are not congruent, and ∢C = 135° . What size is ∢F ?
Answer and solution
The answer: 45 °
The equal angles ∢B, ∢E face the shorter sides AC, DF . The triangles are not congruent, so the angles facing the longer sides add up to 180°: ∢F = 180° − 135° = 45° .
The converse of Pythagoras’ theorem
For example: The sides of a triangle are 5, 8, 10 long. Does the triangle have a right angle?
- Yes
- No
Answer and solution
The answer: No
Check the two shorter sides against the longest: 5² + 8² = 25 + 64 = 89 , and 10² = 100 . They differ, so the triangle has no right angle. No.
The circle round a right triangle
For example: A right triangle has a hypotenuse 8 cm long. What is the radius of its circumscribed circle?
Answer and solution
The answer: 4 cm
The radius is half the hypotenuse: R = 8 ÷ 2 = 4 . The radius is 4 cm.
Not on the site yet
Parts of this topic that are not built yet: The proof of the fourth congruence theorem · Proofs using congruence.
The exercises are made by rules and the answers are worked out in code, so there are new exercises every time. The topic follows the Israeli Ministry of Education’s high school maths curriculum at 5 units (grades 10 to 12).