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 All congruent triangles are similar, but all similar triangles are not congruent.

True. To define similarity simply, it is a case in which all the corresponding angles of two or more triangles are equal, but their sides are not necessarily equal. Congruent triangles are triangles which are identical in every way. As such, all congruent triangles are similar, but all similar triangles are not congruent.

 In a triangle ABC, a line is drawn parallel to BC, which cuts AB at D and AC at E. The ratio of AD:AB is equal to 3:5. What is the ratio of DE:BC?
3:5. Triangle ADE was similar to triangle ABC. A striking property of similar triangles is that the ratio of all their corresponding sides are equal. For example, if triangle ABC is similar to triangle PQR, then:- AB:PQ = AC:PR = BC:QR

 What is the sign which denotes similarity?
~. ABC ~ PQR means that triangle ABC is similar to triangle PQR.

 In a triangle XYZ, a line 'PQ' is drawn parallel to YZ, cutting XY at P and XZ at Q. The ratio of XP:XY is 2:3. What is the ratio XQ:QZ?
2:1. In order to evaluate this answer, you had to use the Basic Proportionality Theorem, which states, "If a line is drawn parallel to a side of a triangle, it divides the other two sides in proportion." For example, in the above-mentioned case, XP:PY = XQ:QZ. Therefore, since the ratio XP:XY was given, you were required to find out the ratio of XP:PY. --------------------------------------------------------------------- XP:XY = 2:3 Let XP = 2x and XY be 3x. PY = XY - XP = 3x - 2x = x. Therefore, XP:PY = 2x:x = 2:1 But XP:PY = XQ:QZ Therefore, XQ:QZ = 2:1 --------------------------------------------------------------------- This was why I advised you to draw diagrams.

 If a line divides two sides of a triangle in proportion, then it is parallel to the third side.
True. What I just stated is the converse of the Basic Proportionality Theorem. --------------------------------------------------------------------- In ABC, DE is a line which cuts AB at D and AC at E. AD:DB = AE:EC. Therefore, DE is parallel to the third side of ABC, namely BC. ---------------------------------------------------------------------

 The Midpoint Theorem states: "A line joining the midpoints of any two sides of a triangle is parallel to the third side and equal to half the length of the third side." So, riddle me this: What is the converse of the Midpoint Theorem?
"If a line passes through the midpoint of one side of a triangle and is parallel to a second side, the given line bisects the third side.". Since I have given you the theorem and its converse, use it to solve any subsequent numerical based on them. Now, I know this is a quiz on Similarity, but I've included the Midpoint and Basic Proportionality Theorems because: 1) The Midpoint Theorem is a special case of similarity. If a line DE passes through the midpoints of the sides AB and AC of a triangle ABC, then triangle ADE automatically become similar to triangle ABC. 2) The Basic Proportionality Theorem is a consequence of similarity. It cannot be proven without similarity.

 Enough about sides. Let's talk areas. You have been given two similar triangles, MNO and DEF. You have to find out the ratios of their areas. Which of the following formulae can be used to find out the ratio of the areas of the given triangles.
The ratio of the areas of the triangles is equal to the square of the corresponding sides. Let me explain a bit more illustratively. [Note: 'Ao' means 'Area of' and '^2' means 'squared'.) --------------------------------------------------------------------- In triangles MNO and DEF:- MN and DE are corresponding sides, MP and DG are corresponding altitudes, and MQ and DH are corresponding medians of triangles MNO and DEF respectively. Ao MNO: Ao DEF = (MN:DE)^2 = (MP:DG)^2 = (MQ:DH)^2 --------------------------------------------------------------------- I'd give you the proof which shows you how to arrive to this conclusion, but it's really long and tedious.

 All equilateral triangles are similar.
True. Each angle of a an equilateral triangle is sixty degrees. Hence, the whole lot of them are similar.

 A given triangle ABC is isosceles. Angle B and angle C are the base angles. From B, a perpendicular 'BD' is drawn to cut AC at point D. From C, a perpendicular 'CE' is drawn to cut AB at point E. State which of the following pairs of triangles are similar.
BDC ~ CEB. It becomes much simpler if you draw a diagram like I told you to. --------------------------------------------------------------------- In triangles BDC and CEB: Angle CEB = Angle BDC = 90 degrees Angle DCB = Angle EBC [As base angles of isosceles triangle ABC are equal] The third pair of angles are also equal, due to the hypothesis that the sum of all the angles of a triangle is equal to 180 degrees.

 ABC and PQR are two triangles. If the ratios of all the corresponding sides of the triangles are equal, then ABC ~ PQR.
True. This is known as the S.S.S test of similarity. --------------------------------------------------------------------- AB:PQ = AC:PR = BC:QR Therefore, ABC ~ PQR. ---------------------------------------------------------------------

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