Similarity of triangle theory and questions.

 Similarity of triangles 

Two triangles are called similar if we can get two congruent triangles after en-

larging or compressing the sides of one of them according to an equal ratio. That

is, two triangles are similar means they have a same shape but may have different

sizes.

Criteria for Similarity of Two Triangles

(I) Each pair of corresponding angles are equal (A.A.A.);

(II) All corresponding sides are proportional (S.S.S.);

(III) Two pairs of corresponding sides are proportional, and the included corre-

sponding angles are equal (S.A.S.)

Basic Properties of Two Similar Triangles

(I) For two similar triangles, their corresponding sides, corresponding heights,

corresponding medians, corresponding angle bisectors, corresponding

perimeter are all proportional with the same ratio.

(II) Consider the similarity as a transformation from one triangle to other, then

this transformation keeps many features of a graph unchanged: each inte-

rior angle is unchanged; any two parallel lines are still parallel, the angle

formed by two intersected lines keeps unchanged, and collinear points re-

mained collinear.

(III) For two similar triangles, the ratio of their areas is equal to square of the

ratio of their corresponding sides.

Illustrations 

1)(Angle Bisector Theorem) For any triangle, the angle bisector of

any interior angle must cut the opposite side into two segments, such that their

ratio is equal to the ratio of the two sides of the angle, correspondingly.

2)(Projection Theorem of Right Triangles) In the right triangle ABC,

∠ACB = 90◦

. Then

CD•CD = AD · DB, AC•AC = AD · AB, BC2 = BD · BA.

3)(Theorem on Centroid) For any triangle ABC, its three medians

must intersect at one common point G, and each median is partitioned by G as

two segments with ratio 2 : 1

All of these questions are available in our video with explanation, except question number 2 which is left as an assignment. 

4)(MOSCOW/1972) In 4ABC, AD, BE are medians on BC, AC

respectively. If ∠CAD = ∠CBE = 30◦

, prove that ∆ABC is equilateral. 

Solution 

∵ ∆ADC ∼ ∆BEC (A.A.A.), therefore

∴ AC²= BC²

, AC = BC.

In ∆BEC, ∠BEC = 30◦

, EC =1/2(BC),

∴ ∠BEC = 90◦

, ∠C = 60◦

∴ ∆ABC is equilateral.

5)