3. In triangles BEF and BDF, BE = BD (Radii of the same arc) EF = DF (Arcs of equal radii) BF = BF (Common) Therefore, BEF = BDF (SSS rule) This gives EBF = DBF (CPCT)
4. CONSTRUCTION OF THE BISECTOR OF A GIVEN ANGLE Bisecting an angle means drawing a ray in the interior of the angle, with its initial point at the vertex of the angle such that it divides the angle into two equal parts. In order to draw a ray AX bisecting a given angle BAC, we following steps. C Q X R A P B Steps of construction STEPI With centre A and any convenient radius draw an are cutting AB and AC at P and Q respectively. STEPII with centre P and radius more than ½ (PQ) draw an arc.
5. STEP III W ith centre Q and the same radius, as in step II, draw another arc intersecting the arc in step II at R. STEPIV Join AR and produce it to any point X. The ray AX is the required bisector of BAC. Verification : Measure BAX and CAX. You would find that BAX = CAX. Justification : Now let us see how this method gives us the required angle bisector: Join PR and QR. In triangles : APR and AQR, we have [ AP and AQ are radii of the same arc ] AP = AQ [PR and QR are arcs of equal radii] PR = QR [Common] AR = AR So, by SSS congruence criterion, we have APR = AQR PAR = QAR Hence, AR is the bisector of BAC.
6. CONSTRUCTION OF SOME STANDARD ANGLES In this section, we will learn how to construct angles of 60 o ,30 0 ,90 0 ,45 0 and 120 0 with the help of ruler and compasses only. For Example : CONSTRUCTION OF AN ANGLE OF 60 0 In order to construct an angle of 60 0 with the help of ruler and compasses only, we follow the following steps. Steps of construction STEPI Draw a ray OA. STEPII With centre O and any radius draw an arc PQ with the help of compasses, cutting the ray OA at P. STEPIII With centre P and the same radius draw an arc cutting the arc PQ at R. STEPIV Join OR and produce it to obtain ray OB.
7. The angle AOB so obtained is the angle of measure 60 0 . Join PR. Justification : Now, let us see how this method gives us the required angle of 60 0. Join PR. In OPR, we have OP = OR = PR [ See construction of angle of 600] OPR is an equilateral triangle. POR = 60 0 [ POR = AOB] AOB = 60 0
8. SOME CONSTUCTIONS OF TRIANGLES In order to construct a triangle at least three parts must be given. But, all the combinations of three parts out of six parts are not sufficient to construct a triangle. For example, if two sides and an angle (not the included angles) are given, then it is not possible to construct such a triangle. CONSTRUCTION OF AN EQUILATERAL TRIANGLE In order to construct an equilateral triangle when the measure (length) of its side is given we follow the following steps: steps of construction STEP I Draw a ray AX with initial point A. STEP II With centre A and radius equal to length of a side of the triangle draw an arc BY, cutting the ray AX at B.
9. STEP III With centre B and the same radius draw an arc cutting the arc BY at C. STEP IV Join AC and BC to obtain the required triangle. CONSTRUCTION OF A TRIANGLE WHEN ITS BASE, SUM OF THE OTHER TWO SIDES AND ONE BASE ANGLE ARE GIVEN In order to construct a triangle, when its base, sum of the other two sides and one of the base angles are given, we follow the following steps: CONSTRUCTIONS Steps of construction STEPI Obtain the base, base angle and the sum of other two sides. Let AB be the base, A be the base angle and I be the sum of the lengths of other two sides BC and CA of ABC. STEPII Draw the base AB.
10. STEPIII Draw BAX of measure equal to that of A STEPIV From ray AX, cut – off line segment AD equal to 1 (the sum of other two sides). STEPV Join BD. STEPVI Draw the perpendicular bisector of BD meeting AD at C. STEPVII Join BC to obtain the required triangle ABC. Justification : Let us now see how do we get the required triangle: since point Clies on the perpendicular bisector of BD. Therefore, CD = CB Now AC = AD – CD AC = AD – CB [CD = CB] AD = AC + CB
11. CONSTRUCTION OF A TRIANGLE WHEN ITS BASE, DIFFERENCE OF THE OTHER TWO SIDES AND ONE BASE ANGLE ARE GIVEN In order to construct a triangle when its base, difference of the other two sides and one of the base angles are given, we follow the following steps:
12. Steps of construction STEPI Obtain the base, base angle and the difference of two other sides. Let AB be the base, A be the base angle and I be the difference of the other two sides BC and CA of ABC. i.e., I= AC – BC, if AC >BC or, I= BC – AC, if BC >AC STEPII Draw the base AB of given length. STEPIII Draw < BAX of measure equal to that of <A STEPIV If AC>BC, then cut off segment AD = AC – BC from ray AX. (See fig 16.18.(i)) if AC < BC then extend XA to X’ on opposite side of AB and cut off segment AD = BC – AC from ray AX’. (See fig. 16.18 (ii)). STEPV Join BD.
13. STEPVI Draw the perpendicular bisector of BD which cuts AX or AX’, as the case may be, at C. STEPVII Join BC to obtain the required triangle ABC. Justification: Let us now see how do we get the required triangle. Since C lies on the perpendicular bisector of DB. So, CD = CB AD = AC – CD = AC - BC
14. CONSTRUCTION OF A TRIANGLE OF GIVEN PERIMETER AND TWO BASE ANGLES In order to construct a triangle of given perimeter and two base angles, we follow the following steps: Steps of construction STEPI Obtain the perimeter and the base angles of the triangle. Let ABC be a triangle of perimeter p cm and base BC.
15. STEPII Draw a line segment XY equal to the perimeter p of ABC. STEPIII Construct YDX = B and XYE = C. STEPIV Draw bsectors of angles < YXD and XYE and mark their intersection point as A. STEPV Draw the perpendicular bisectors of XA and YA meeting XY in B and C respectively. STEPVI Join AB and AC to obtain the required triangle ABC. Justification : For the justification of the construction, we observe that B lies on the perpendicular bisector of AX. XB = AB < AXB = BAX Similarly, Clies on the perpendicular bisector of AX.
16. YC = AC AYC = YAC Now, XY = XB + BC + CY XY = AB + BC + AC In AXB, we have ABC = AXB + BAX = 2 AXB = BXD = BXY = B. In AYC, we have ACB = AYC + YAC = 2 AYC = CYE = C.
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