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.

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

Solution:

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

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

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

  • Draw bisectors of ∠B and ∠ C .

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

  • Draw IK BC

  • Also draw IJ AB and IL AC.

  • Join AI.

  • In BIK and BIJ ,

    ∠IKB = ∠IJB = 90° [By construction]

    ∠IBK = ∠IBJ

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

    BI=BI [Common]

    BIKBIJ [Angle-Side-Angle criteria of congruency]

    IK=IJ [By Corresponding Parts Congruent Triangles.] ……….equation(1)

  • Similarly, CIK CIL

    IK=IL [By Corresponding Parts Congruent Triangles.] ……….equation(2)

  • From eq (1) and (2),

    IK=IJ=IL

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

  • Draw bisectors of B and C .

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

  • Draw IKBC

  • Also draw IJAB and ILAC .

  • Join AI.

  • In BIKandBIJ,

    IKB=IJB=90° [By construction]

    IBK=IBJ

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

    BI=BI [Common]

    BIKBIJ [Angle-Side-Angle criteria of congruency]

    IK=IJ [By Corresponding Parts of Congruent Triangles.] ……….equation(1)

  • Similarly ,CIKCIL

    IK=IL [By Corresponding Parts Congruent Triangles.] ……….equation(2)

  • From eq (1) and (2),

    IK=IJ=IL

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

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