NCERT Class 9 Solutions: Triangles (Chapter 7) Exercise 7.5 – Part 2

Perpendicular bisector

Give perpendicular bisector at triangle ABC ,D is a mid point of AB,E is a mid point of BC and F is mid point of AC

Give Perpendicular Bisector at Triangle ABC

Give perpendicular bisector at triangle ABC ,D is a mid point of AB,E is a mid point of BC and F is mid point of AC

To find this point, you will construct three perpendicular bisectors, one for each side of the triangle. The point where all three perpendicular bisectors intersect is called the Circumcenter. Using this center point, we can draw a circle that passes through all three vertices.

Q-2. In the triangle located a point in its interior which is equidistant from all the side of the triangle.

Solution:

Angle α Angle α: Angle between F, G, F' Angle β Angle β: Angle between B, D, B' Angle γ Angle γ: Angle between F, E, F'_1 Segment f Segment f: Segment [A, B] Segment g Segment g: Segment [B, C] Segment h Segment h: Segment [A, C] Segment i Segment i: Segment [C, D] Segment j Segment j: Segment [B, E] Segment k Segment k: Segment [F, G] A text1 = "A" B text2 = "B" C text3 = "C" J text4 = "J" K text5 = "K" I text6 = "I" L text7 = "L"

Triangle of ABC Its Bisectors of ∠B and ∠C.

Triangle of ABC its bisectors of ∠B and ∠C., this angle bisectors intersect each other at point l

  • Draw bisectors of ∠B and ∠ Equation .

  • Let these angle bisectors intersect each other at point I.

  • Draw IK Equation BC

  • Also draw IJ Equation AB and IL Equation AC.

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  • In Equation and Equation ,

    ∠IKB Equation ∠IJB = Equation [By construction]

    ∠IBK Equation ∠IBJ

    [ Equation BI is the bisector of ∠B (By construction)]

    Equation [Common]

    Equation Equation [Angle-Side-Angle criteria of congruency]

    Equation [By Corresponding Parts Congruent Triangles.] ……….equation(1)

  • Similarly, Equation CIK Equation CIL

    Equation [By Corresponding Parts Congruent Triangles.] ……….equation(2)

  • From eq (1) and (2),

    Equation

  • Hence, I is the point of intersection of angle bisectors of any two angles of Equation equidistant from its sides.

  • Draw bisectors of Equation and Equation .

  • Let these angle bisectors intersect each other at point I.

  • Draw Equation

  • Also draw Equation and Equation .

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  • In Equation

    Equation [By construction]

    Equation

    [ Equation BI is the bisector of Equation B (By construction)]

    Equation [Common]

    Equation [Angle-Side-Angle criteria of congruency]

    Equation [By Corresponding Parts of Congruent Triangles.] ……….equation(1)

  • Similarly Equation

    Equation [By Corresponding Parts Congruent Triangles.] ……….equation(2)

  • From eq (1) and (2),

    Equation

  • Hence, I is the point of intersection of angle bisectors of any two angles of Equation equidistant from its sides.

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