NCERT Class 9 Solutions: Triangles (Chapter 7) Exercise 7.1 – Part 3

Angle Angle Side congruence condition

Give Angle Angle Side congruence condition of ΔENR and ΔRNV

Give Angle Angle Side Congruence Condition

Give Angle Angle Side congruence condition of ΔENR and ΔRNV

5. Line l is the bisector of an angle A and B is any point on l. BP and BQ are perpendiculars from B to the arms of A (see Fig). Show that:

  1. ΔAPBΔAQB

  2. BP=BQ or B is equidistant from the arms of A.

Give Angle a and B is any point on l that line is bisector.BP and BQ are perpendicular from B tohe arms of angle A

Give Angle a and B Is Any Point on L

Give Angle a and B is any point on l that line is bisector.BP and BQ are perpendicular from B tohe arms of angle A

Solution:

Give, l is the bisector of an angle A.

BP and BQ are perpendiculars.

  1. In ΔAPB and ΔAQB,

    P=Q (Right angles)

    BAP=BAQ (l is bisector)

    AB=AB (Common line)

    Therefore, ΔAPBΔAQB (by Angle-Angle-Side congruence condition).

  2. BP=BQ by Corresponding Part of Congruent Triangles.

    Therefore, B is equidistant from the arms of A .

Q-6 In the Figure. , AC=AE,AB=ADandBAD=EAC . Show that BC=DE.

 Give point of A,B,C,D,E ,∠BAD = ∠EAC also AC=AE,AB=AD

Give Point of a,B,C,D,E ,∠BAD = ∠EAC

Give point of A,B,C,D,E ,∠BAD = ∠EAC also AC=AE,AB=AD

Solution:

Given,

AC=AE,

AB=ADandBAD=EAC

To show,

BC=DE

Proof,

BAD=EAC (Adding ∠DAC both sides)

BAD+DAC=EAC+DAC

BAC=EAD

In ΔABC and ΔADE,

AC=AE (Given)

BAC=EAD

AB=AD (Given)

Therefore, ΔABCΔADE (by Side-Angle-Side congruence condition).

BC=DE (By Corresponding Part of Congruent Triangles)

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