Lessons 12–14 · 3 lessons · B. Xaydarov, E. Sariqov, A. Qo‘chqorov. Geometry, Grade 9 (textbook for general secondary schools), revised 4th edition. Huquq va Jamiyat Publishing, Tashkent, 2019
12
Using similarity tests in proof problems
Textbook: pp. 40–41
GoalApply the similarity tests in proof problems; know the property that an angle bisector divides the opposite side proportionally.
Similarity tests are a powerful way to get an equality or relation between lengths in proof problems. Usually we find the two relevant triangles, justify their similarity by AA, SAS or SSS, and then write an equality from the ratios of corresponding sides. An important result: the bisector of a triangle's angle divides the opposite side into segments proportional to the adjacent sides: if AD is the bisector in triangle ABC, then BD : DC = AB : AC. In the proof an extra parallel line is drawn to create similar triangles. It is advisable to write a solution in the order “given – to prove – proof”.
Worked examples
In triangle ABC, AD is the bisector, AB = 6, AC = 9, BC = 10. BD : DC = 6 : 9 = 2 : 3, so BD = 10 · 2/5 = 4 and DC = 6.
Midline MN of a triangle: △BMN ∽ △BAC (common ∠B, BM/BA = BN/BC = 1/2), so MN = AC/2 and MN ∥ AC.
Class activity
“Rebuilding the proof”: the teacher writes the steps of a proof on cards in mixed order; groups put them in the right order and give the reason for each step.
Practice
1
AD is the bisector, AB = 6, AC = 10, BC = 8. How long is BD?
3
2
AD is a bisector, BD = 5, DC = 10 and AB = 7. How long is AC?
14
3
If △BMN ∽ △BAC with k = 1/2, what is the ratio of the areas?
1/4
4
Why does the bisector property give segments in the same ratio as the adjacent sides?
It follows from the proportional corresponding sides of similar triangles built with an extra parallel line.