Triangles. Example: In the given figure, S and T are points on PQ and PR respectively of PQR such that ST QR. Determine the length of PR.


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1 Triangles Two geometric figures having the same shape and size are said to be congruent figures. Two geometric figures having the same shape, but not necessarily the same size, are called similar figures. Example: All circles are similar. All equilateral triangles are similar. All congruent figures are similar. However, the converse is not true. Two polygons with the same number of sides are similar, if their corresponding angles are equal their corresponding sides are in the same ratio (or proportion) Similarity of Triangles Two triangles are similar, if their corresponding angles are equal their corresponding sides are in the same ratio (or proportion) Theorem 1: (Basic proportionality theorem) If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, then the other two sides are divided in the same ratio. Corollary: If D and E are points on the sides, AB and AC, respectively of ABC such that DE BC, then AB AC AD AE AB AC DB EC Theorem 2: (Converse of basic proportionality theorem) If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side. Example: In the given figure, S and T are points on PQ and PR respectively of PQR such that ST QR. Determine the length of PR.
2 Solution: Since ST QR, by basic proportionality theorem, we have PS PT SQ TR TR 31.8 TR 4.5 cm 1.2 PR PT TR cm 7.5 cm Example: In the given figure, AB DE and AD EF. Show that BD:DC::DF:FC. Solution: In ABC, AB DE BD AE DC EC (1) [By BPT] In ADC, AD EF AE DF (2) [By BPT] EC FC From (1) and (2), we obtain BD DF DC FC BD:DC::DF:FC Criteria for similarity of triangles
3 Theorem 3: (AAA similarity criterion) If in two triangles, corresponding angles are equal, then their corresponding sides are in the same ratio (or proportion) and hence, the two triangles are similar. Theorem 4: (AA similarity criterion) If in two triangles, two angles of one triangle are respectively equal to the two angles of the other triangle, then the two triangles are similar. Theorem 5: (SSS similarity criterion) If in two triangles, sides of one triangle are proportional to the sides of the other triangle, then their corresponding angles are equal and hence, the two triangles are similar. Theorem 6: (SAS similarity criterion) If one angle of a triangle is equal to one angle of the other triangle and the sides including these angles are proportional, then the two triangles are similar. Example: In ABC, C is acute, D and E are points on sides BC and AC respectively, such that ADBC and BEAC. Show that BC CD = AC CE Solution: In ADC and BEC, ADC = BEC = 90 DCA = ECB [Common] By AA similarity criterion, ADC BEC CD AC CE BC BC CD AC CE Hence, the result is proved. Areas of similar triangles Theorem 7: The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.
4 Example: In ABC, D and E are the respective midpoints of sides AB and BC. Find the ratio of the areas of DBE and ABC. Solution: In ABC, D and E are the respective midpoints of the sides, AB and BC. By the converse of BPT, DE AC In DBE and ABC, DBE = ABC BED = BCA BDE = BAC [Common] [Corresponding angles] [Corresponding angles] By AAA similarity criterion, DBE ABC Area DBE BE Area ABC BC 2 BE 2BE 1 4 1: 4 2 E is midpoint of BC Result: Using the above theorem, the result the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding medians or altitudes or angle bisector can be proved. Theorem 8: If a perpendicular is drawn from the vertex of the right angle of a right triangle to the hypotenuse, then triangles on both sides of the perpendicular are similar to the whole triangle and to each other. Theorem 9: (Pythagoras theorem) In a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.
5 Result: In any triangle, three times the sum of the squares of the sides of a triangle is equal to four times the sum of the squares of the medians of the triangle. Theorem 10: (Converse of Pythagoras theorem) In a triangle, if the square of one side is equal to the sum of the squares of other two sides, then the angle opposite to the first side is a right angle. Example: ABC is rightangled at B, BDCA. Prove that BD 2 = CD DA Solution: By applying Pythagoras theorem in BDC, BDA, and ABC, we obtain BC 2 = CD 2 + BD 2 (1) BA 2 = BD 2 + DA 2 (2) CA 2 = BC 2 + BA 2 (3) Adding equations (1) and (2), we obtain BC 2 + BA 2 = 2BD 2 + CD 2 + DA 2 CA 2 = 2BD 2 + CD 2 + DA 2 [Using (3)] (CD + DA) 2 = 2BD 2 + CD 2 + DA 2 CD 2 + DA CD DA = 2BD 2 + CD 2 + DA 2 CD DA = BD 2 Hence, the result is proved.
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