GEOMETRY. Similar Triangles

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1 GOMTRY Similar Triangles SIMILR TRINGLS N THIR PROPRTIS efinition Two triangles are said to be similar if: (i) Their corresponding angles are equal, and (ii) Their corresponding sides are proportional. It follows from this definition that two triangles and F are similar if: (i) =, =, = F, and (ii). F F Note: In the later part of this chapter we shall show that the two conditions given in the above definition are not independent. In fact, if either of two conditions holds, then the other holds automatically. So, any one of the two conditions can be used to define similar triangles. SOM SI RSULTS ON PROPORTIONLITY In this section we shall discuss some basic results on proportionality. Theorem 1: orollary (asic proportionality Theorem or Thales Theorem) If a line is drawn parallel to one side of a triangle intersecting the other two sides, then it divides the two sides in the same ratio. If in a, a line, intersects in and in, then: (i) (ii). Theorem : (onverse of asic Proportionality Theorem) If a line divides any two sides of a triangle in the same ratio, then the line must be parallel to the third side. Illustration 1: Prove Pythagoras theorem by means of similar triangles. Solution: Let be a triangle right angled at. Let be perpendicular to. From triangle, = 90 = 90; = 90 = 90 (90 ) = Triangle and are equiangular and hence similar. i.e., ross multiplying =. dding (1) and () + =. +. = ( + ) =. = (Since + = ) i.e., sum of the squares on the sides containing the right angle is equal to square on the hypotenuse, which is Pythagoras theorem.

2 Illustration : The areas of two similar triangle and PQR and 64cm and 11cm respectively. If QR = 15.4 Solution: cm, find. The two triangles and PQR are similar. Hence the ratio of the areas of them is equal to the ratio of the squares of their corresponding sides. P PQR QR (15.4) 8 8 = (15.4) = (8 1.4) = = 11. cm Q R Illustration 3: is a trapezium in which and =. Find the ratio of the areas of triangles O Solution: and O. onstruction: Join and, In two triangles O and O, = O = O O = O O = O, (vertically opposite angles.) the two triangles are equiangular and hence similar. O 1 O 4 4 O 4 O ( = ) Illustration 4: XY and XY divides triangular region into two parts equal in area. etermine X. Solution: XY is drawn parallel to. In two triangles XY and, is ommon XY = YX = XY and are equiangular and hence similar. XY X Now the line XY divides the triangular region into two parts whose areas are equal. If =. X 1 X 1 X X 1 X X 1 X O Y xercise 1: In figure. If = x, = x, = x + and = x 1, find the value of x.

3 xercise : Let X be any point on the side of a triangle. If XM, XN are drawn parallel to and meeting, in M, N respectively; MN meets produced in T, prove that TX = T T. xercise 3: is a parallelogram, P is a point on side and P when produced meets produced at L. Prove that (i) P = (ii) L = L PL L P xercise 4: is a quadrilateral; P, Q, R and S are the points of trisection of sides,, and respectively and are adjacent to and ; prove that PQRS is a parallelogram. xercise 5: The side of a triangle is bisected at ; O is any point in. O and O produced meet and in and F respectively and is produced to X so that is the mid-point of OX. Prove that O : X = F : and show that F. More Results ased on asic Proportionality Theorem and its onverse Theorem 3: The internal bisector of an angle of a triangle divides the opposite side internally in the ratio of the sides containing the angle. Theorem 4: In a triangle, if is a point on such that, prove that is the bisector of. Theorem 5: The external bisector of an angle of a triangle divides the opposite side externally in the ratio of the sides containing the angle. xercise 6: In figure is the bisector of. If = 10cm, = 14 cm and = 6 cm, find : cm 10 cm x 6 - x xercise 7: The bisector of interior of meets in, and the bisector of exterior meets produced in. Prove that =. xercise 8: is a quadrilateral in which =. The bisector of and intersect sides and at the points and F respectively. Prove that F. xercise 9: O is any point inside a triangle. The bisectors of O, O and O meets the sides, and in point, and F respectively. Show that F = F. xercise 10: Theorem 6: is a median of. The bisectors of and meets and in and F respectively. Prove that F. The line drawn from the mid-point of one side of a triangle parallel to another side bisects the third side.

4 Theorem 7: Theorem 8: Theorem 9: Theorem 10: Theorem 11: OROLLRY The line joining the mid-points of two sides of a triangle is parallel to the third side and half of the third side. The diagonals of a trapezium divide each other proportionally. If the diagonals of a quadrilateral divide each other proportionally, then it is a trapezium. ny line parallel to the parallel sides of a trapezium divides the non-parallel side proportionally. If three or more parallel lines are intersected by two transversals, prove that the intercepts made by them on the transversals are proportional. If three or more parallel straight lines make equal intercepts on a given transversal, prove that they will make equal intercepts on any other transversal. haracteristic Properties of Similarity We have defined similarity of two triangles. Two triangles are said to be similar iff (i) their corresponding angles are equal and (ii) their corresponding sides are proportional. Thus, two triangles and F are similar if. (i) =, =, = F, (ii) F F quiangular Triangles Two triangles are said to be equiangular, if their corresponding angles are equal Theorem 1: ( Similarity) If two triangles are equiangular, then the triangles are similar. (orresponding sides are those, which are opposite to equal angles) G H F ata: Let, F be two triangles, such that = ; = and = F To Prove: F F Proof: ut off G = and and H = from F, Join GH. In s and GH = GH = G = H = GH = GH ut = F (by data) GH = F ut there are corresponding angles. GH is parallel to F. G H F G = and H =

5 F Similarly it can be proved that F F F If in two triangles and F if = ; = and = F then F F i.e., if two triangles are equiangular, then their corresponding sides are proportional i.e., if a, b, c are the length of the sides of the triangle and d, e, f are the lengths of the triangle F then a b c k(a constant), d e f Then a = kd; b = ke; c = kf. The converse of the above theorem is given below. Theorem 13: similar. (SSS Similarity) If the corresponding sides of two triangles are proportional, then they are F G ata: Let and F be two triangles such that F F To Prove: = ; = ; = F onstruction: On side of F opposite to, Make FG = and FG = Proof: In triangles and GF FG = and FG = GF = the two triangles and FG are equiangular and hence their corresponding G F ut (y data) F G =. Similarly it can be proved that FG = F In the triangles F and GF, F is common = F F = FG Hence the s F and GF are congruent F = GF = (by construction) and F = FG =

6 Triangles and F are equangular Theorem 14: (SS Similarity) If in two triangles, one pair of corresponding sides are proportional and the included angles are equal then the two triangles are similar. R G H F ata : Let and F be two triangles in which = and F To Prove: Triangles and F are equiangular, onstruction: long, cut off G = and along F cut off H =, Join GH Proof: In triangles and GH = G = H = the two triangles and GH are congruent GH = (ata) F i.e., G H F Hence GH is parallel to F GH = corresponding ut GH = = lso = (data) = F Hence the triangle, F are equiangular. Remarks: If two triangles and F are similar, then F F Theorem 15: If two triangles are equiangular, prove that the ratio of the corresponding sides is same as the ratio of the corresponding angle bisector segments. Theorem 16: If two triangles are equiangular, prove that the ratio of the corresponding sides is same as the ratio of the corresponding attitudes. Theorem 17: If one angle of a triangle is equal to one angle of another triangle and the bisector of these equal angles divides the opposite side in the same ratio, prove that the triangles are similar. Theorem 18: If two sides and a median bisecting one of these sides of a triangle are respectively proportional to the two sides and the corresponding median of another triangle, then the triangles are similar.

7 Theorem 19: If two sides and a median bisecting the third side of a triangle are respectively proportional to the corresponding sides and the median of another triangle, then the two triangles are similar. Theorem 0: The ratio of the area of two similar triangles is equal to the ratio of the squares of any two corresponding sides. Theorem 1: The areas of two similar triangles are in the ratio of the squares of the corresponding altitudes. Theorem : The areas of two similar triangles are in the ratio of the squares of the corresponding medians. Theorem 3: The areas of two similar triangles are in the ratio of the squares of the corresponding angle bisector segments. Theorem 4: If the areas of two similar triangles are equal, then the triangles are congruent i.e. equal and similar triangles are congruent. Illustration 5: is a parallelogram. The side is bisected at and meets at F. Prove that F = /3. Solution: In triangles F and F F = F (vertically opposite angles) F = alt. F third angles are equal. Hence the two triangles are equiangular and hence their corresponding sides are proportional. F F 1 F F F 1 F F F 1 3 F i.e., or F 3 3 Illustration 6: Two sides of a triangle are 10m and 15m long, and the base is 0m long. If another triangle similar Solution: to the first, has the base of length 3.5m, then find the lengths of the other sides. Let k be the ratio of the corresponding sides k = The length of the other sides of the second triangle are 10k and 15k i.e., 10 = 16.5m and m xercise 11: Prove that the area of the triangle described on one side of a square as base is one half the area of the similar triangle F described on the diagonal as base. xercise 1: Prove that the area of the equilateral triangle described on the side of a square is half the area of the equilateral triangle described on its diagonal.

8 xercise 13: In figure is a trapezium in which and =. etermine the ratio of the areas of O and O. o xercise 14: In figure P, Q and R are each perpendicular to. Prove that = x z y. x P Q R z xercise 15: In figure, if =, prove that -. lso, find the value of x x 9 xercise 16: In figure, P is the mid-point of and Q is the mid-point of P. If Q when produced meets at R, prove that RS = ⅓. Q R S P xercise 17: The diagonal of a parallelogram intersects the segment at the point F, where is any point on the side. Prove that F F = F F. 4 F 1 3 xercise 18: Two isosceles triangles have equal vertical angles and their areas are in the ratio 16 : 5. Find the ratio of their corresponding heights. Pythagoras Theorem Statement In a right angled triangle, the square of the hypotenuse is equal to the sum of the squares of the order two sides. onverse of Pythagoras Theorem In a triangle, if the square of one side is equal to the sum of the squares of the other two sides, then the angle opposite to the first side is a right angle.

9 Theorem 1: (RSULT ON OTUS TRINGL) In the given figure, is an obtuse triangle, obtuse-angled at. If, Prove that = Remark In the above theorem is known as the projection of on and the theorem can also be stated as: In another obtuse triangle, the square of the side opposite to obtuse angle is equal to the sum of the squares of other two sides plus twice the product of one side and the projection of other on first. Theorem : (RSULT ON UT TRINGL) In the given figure, of is an acute angle and, prove that = +. Remark In the above theorem is known as the projection of on and the theorem can also be stated as: In an acute triangle, the square of the side opposite to an acute angle is equal to the sum of the squares of other two sides minus twice the product of one side and the projection of other on first. Theorem 3: Theorem 4: Prove that in any triangle, the sum of the squares of any two sides is equal to twice the square of half of the third side together with twice the square of the median which bisects the third side. Prove that three times the sum of the squares of the side of a triangle is equal to four times the sum of the squares of the medians of the triangle. Illustration 7: is a right triangle, right angled at. Let and be any points on and respectively. Prove that + = + Solution: Since is right triangle, right-angled at. = +..(i) again, is right triangle, right-angled at. = +..(ii) dding (i) and (ii), we get + = ( + ) + ( + ) = ( + ) + ( + ) = + [Using Pythagoras theorem for and, we have = + + ] Hence, + = +. Theorem 5: The perpendicular on the base of a intersects at so that = 3. Prove that = +.

10 Illustration 8: is a right triangle, right-angled at. Let = a, = b, = c and let p be the length of perpendicular from on, prove that 1 1 (i) cp = ab (ii) + 1 p a b Solution: Let. Then, = p. c p b area of = 1 (ase height) = 1 ( ) = 1 cp. a lso, area of = 1 ( ) = 1 ab 1 cp = 1 ab cp = ab (ii) Since is a right triangle, right angled at. = + c = a + b ab p ab p = a + b = a + b a b 1 = p ab 1 = p 1 b + 1 a 1 p = 1 a + 1 b Illustration 9: Solution: Prove that three times the square of any side of an equilateral triangle is equal to four times the square of the altitude. Let be an equilateral triangle and let. In, we have: = (given) = and = [ach equal to 90 0 ]. So, = = = 1. Since, is a right triangle, right-angled at. = + = 1 + = + 4 = = 3 = 4. xercise 19: In figure M M; determine M in terms of x, y and z. x M y z

11 xercise 0: In figure. Prove that (i) MU MV (ii) M V = M U U M v xercise 1: orresponding sides of two triangles are in the ratio : 3. If the area of the smaller triangle in 48 cm, determine the area of the large triangle. xercise : In, ray bisects and intersects in. If = a, = b and =c, prove that (i) = ac b + c (ii) = ab b + c xercise 3: In, is an obtuse angle. and = + 3. Prove that =. xercise 4: point is an the side of an equilateral triangle such that = 1 4 = 13.. Prove that xercise 5: Prove that in any triangle, the sum of the squares of any two sides is equal to twice the square of half of the third side together with twice the square of the median which bisects the third side. xercise 6: Prove that in an equilateral triangle, three times the square of a side is equal to four times the square of its altitudes. xercise 7: In a triangle, N is a point on such that N. If N = N. N, prove that = nswers to exercise: 1. x = 4 6. = 5/, = 7/ 15. x = M = yz/x m. sq.

12 SSIGNMNTS SUJTIV LVL I 1. Show that any point on the bisector of an angle is equidistant from the arms of the angle.. Show that the straight lines which join the extremities of the base of on isosceles triangle to the middle points of the opposite sides are equal to one another. 3. In the triangle, the bisectors of exterior angles and meet at O. Given that = 70, = 50. Find O If two straight lines are perpendicular to two other straight lines each to each, then show that the acute angle between the first pair is equal to the acute angle between the second pair. O 5. In a quadrilateral, if = and =, then show that the diagonal bisects the angles, which it joins and that is perpendicular to. 6. (i) Show that in a triangle, any two sides are together greater than twice the median, which bisects the remaining side. (ii) Show that in any triangle, the sum of the medians is less than the perimeter. 7. mong all straight line segments drawn from a given point to a given straight line, show that the perpendicular is the least. 8. If the opposite sides of a quadrilateral are equal, then show that the figure is a parallelogram. 9. Show that the line segment joining the meddle points of two sides of a triangle is parallel to the third side and is equal to half the third side. 10. Show that the straight line drawn through the middle point of a side of a triangle, parallel to the base, bisects the remaining side.

13 LVL II 1. In, the bisector X of intersects at X. XL r and XM r are drawn. Prove that XM = XL.. P and Q are two isosceles triangle on same base but on opposite side of line. Show that the line joining P and Q bisect at right angle. 3. In the given fig. PR = QR. SR = TR. N, M are mid points of SR and TR respectively. PRT = QRS. Prove that QM = PN. P R N S M Q T 4. In the, the sides, are produced and bisectors of the exterior angle and are drawn to meet in I 1. Prove that I 1 = 90 /. 5. In a quadrilateral. If = and =, then prove that the figure in a parallelogram.

14 OJTIV LVL I 1. The area of equilateral triangle with side a is () 3 a. () () a 3 3 a. Then its altitude is 4 () 3a a 3. In equilateral, is altitude. Then 4 =.. () () () () 3 3. In an equilateral, is trisected at. Then 7 = () 9 () 9 () 9 () 9 4. In the figure, P and Q are midpoints of and. Then 5 =.. () 4(Q + P ) () 4( + () P + P () P + Q P Q 5. In figure, = P. Then c a =.. () c b () b p () p b () ab pc P a LVL II

15 1. In,, & F are medians. Then 4( + + F ) = () 3(O + O + O ) () 3(O + OF + O ) () 3( + + ) () 3( + F + ) F o. In right angled t, is median. Then = () () () 3 4 () In, and F then =.. () ( F) + ( ) () ( F) + ( ) () ( F) + ( ) () + + F 4. In the figure, is isosceles. Then = () () () () 5. In, > and. Then =.. () () () ()

16 NSWRS SUJTIV LVL I 1. m /n metres 9. = 39, = (a b) a LVL II OJTIV LVL I LVL II

1 st Preparatory. Part (1)

1 st Preparatory. Part (1) Part (1) (1) omplete: 1) The square is a rectangle in which. 2) in a parallelogram in which m ( ) = 60, then m ( ) =. 3) The sum of measures of the angles of the quadrilateral equals. 4) The ray drawn