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

(i) ΔABD ≅ ΔBAC

(ii) BD = AC

(iii) ∠ABD = ∠BAC.

**Answer**

Given,

AD = BC and ∠DAB = ∠CBA

(i) In ΔABD and ΔBAC,

AB = BA (Common)

∠DAB = ∠CBA (Given)

AD = BC (Given)

Therefore, ΔABD ≅ ΔBAC by SAS congruence condition.

(ii) Since, ΔABD ≅ ΔBAC

Therefore BD = AC by CPCT

(iii) Since, ΔABD ≅ ΔBAC

Therefore ∠ABD = ∠BAC by CPCT

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

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

**Answer**

Given,

AD and BC are equal perpendiculars to AB.

To prove,

CD bisects AB

Proof,

In ΔAOD and ΔBOC,

∠A = ∠B (Perpendicular)

∠AOD = ∠BOC (Vertically opposite angles)

AD = BC (Given)

Therefore, ΔAOD ≅ ΔBOC by AAS congruence condition.

Now,

AO = OB (CPCT). CD bisects AB.

l and m are two parallel lines intersected by another pair of parallel lines p and q (see Fig. 7.19). Show that ΔABC ≅ ΔCDA.

l and m are two parallel lines intersected by another pair of parallel lines p and q (see Fig. 7.19). Show that ΔABC ≅ ΔCDA.

**Answer**

Given,

l || m and p || q

To prove,

ΔABC ≅ ΔCDA

Proof,

In ΔABC and ΔCDA,

∠BCA = ∠DAC (Alternate interior angles)

AC = CA (Common)

∠BAC = ∠DCA (Alternate interior angles)

Therefore, ΔABC ≅ ΔCDA by ASA congruence condition.

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. 7.20). Show that:

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. 7.20). Show that:

(i) ΔAPB ≅ ΔAQB

(ii) BP = BQ or B is equidistant from the arms of ∠A.

Answer

Given,

l is the bisector of an angle ∠A.

BP and BQ are perpendiculars.

(i) In ΔAPB and ΔAQB,

∠P = ∠Q (Right angles)

∠BAP = ∠BAQ (l is bisector)

AB = AB (Common)

Therefore, ΔAPB ≅ ΔAQB by AAS congruence condition.

(ii) BP = BQ by CPCT. Therefore, B is equidistant from the arms of ∠A.