NCERT Solutions
Class 9 - Mathematics - Chapter 7: TRIANGLES

EXERCISE 7.1

In quadrilateral ACBD, AC = AD and AB bisects \(\angle A\) (see Fig. 7.16). Show that \(\triangle ABC \cong \triangle ABD\). What can you say about BC and BD?

ABCD is a quadrilateral in which AD = BC and \(\angle DAB = \angle CBA\) (see Fig. 7.17). Prove that:

(i) \(\triangle ABD \cong \triangle BAC\)
(ii) BD = AC
(iii) \(\angle ABD = \angle BAC\)

AD and BC are equal perpendiculars to a line segment AB (see Fig. 7.18). Show that CD bisects AB.

Lines \(l\) and \(m\) are two parallel lines intersected by another pair of parallel lines \(p\) and \(q\) (see Fig. 7.19). Show that \(\triangle ABC \cong \triangle CDA\).

Line \(l\) is the bisector of an angle \(\angle A\) and B is any point on \(l\). BP and BQ are perpendiculars from B to the arms of \(\angle A\) (see Fig. 7.20). Show that:

(i) \(\triangle APB \cong \triangle AQB\)
(ii) BP = BQ or B is equidistant from the arms of \(\angle A\).

In Fig. 7.21, AC = AE, AB = AD and \(\angle BAD = \angle EAC\). Show that BC = DE.

AB is a line segment and P is its mid-point. D and E are points on the same side of AB such that \(\angle BAD = \angle ABE\) and \(\angle EPA = \angle DPB\) (see Fig. 7.22). Show that:

(i) \(\triangle DAP \cong \triangle EBP\)
(ii) AD = BE.

In right triangle ABC, right angled at C, M is the mid-point of hypotenuse AB. C is joined to M and produced to a point D such that DM = CM. D is joined to B (see Fig. 7.23). Show that:

(i) \(\triangle AMC \cong \triangle BMD\)
(ii) \(\angle DBC\) is a right angle.
(iii) \(\triangle DBC \cong \triangle ACB\)
(iv) \(CM = \tfrac{1}{2} AB\).