**TEXTBOOK’S EXERCISE – 7.1 **

**In quadrilateral ACBD, AC = AD and AB bisect ∠ A (see Fig.). Show that ΔABC ≅ ΔABD. What can you say about BC and BD?**

**Sol. **In ΔABC and ΔABD, we have

AC = AD [Given]

∠CAB = ∠DAB [Since AB bisects ∠A]

AB = AB [Common]

∴
ΔABC ≅ ΔABD [By SAS congruence] **Proved. **

Therefore, BC = BD [CPCT].

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

**(i) Δ ABD ≅ Δ BAC (ii) BD = AC (iii) ∠ ABD = ∠ BAC**

**Sol. **In ΔABD and ΔBAC, we have

AD = BC [Given]

∠DAB = ∠CBA [Given]

AB = AB [Common]

∴ ΔABD ≅ ΔBAC [By SAS congruence]

∴ BD = AC [CPCT]

and
∠ABD = ∠BAC [CPCT] **Proved. **

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

**Sol. **In ΔAOD and ΔBOC, we have,

∠AOD = ∠BOC [Vertically opposite angles]

∠CBO = ∠DAO [Each = 90°]

and AD = BC [Given]

∴ ΔAOD ≅ ΔBOC [By AAS congruence]

⇒ AO = BO [CPCT]

Hence, CD bisects AB. **Proved.**

**4. l and m are two parallel lines intersected by another pair of parallel lines p and q (see Fig.). Show that Δ ABC ≅ ΔCDA.**

**Sol. **In the given figure, ABCD is a parallelogram in which
AC is a diagonal, i.e., AB || DC and BC || AD.

In Δ ABC and Δ CDA, we have,

∠BAC = ∠DCA [Alternate angles]

∠BCA = ∠DAC [Alternate angles]

AC = AC [Common]

∴ Δ ABC ≅ Δ CDA

[By ASA congruence] **Proved. **

**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:**

**(i) Δ APB ≅ Δ AQB **

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

**Sol. **In Δ APB and Δ AQB, we have

∠PAB = ∠QAB [Since l is the bisector of ∠ A]

∠APB = ∠AQB [Each = 90°]

AB = AB [Common]

∴ ΔAPB ≅ ΔAQB [By AAS congruence]

⇒ BP = BQ [CPCT]

i.e., B is equidistant from the arms of ∠A. **Proved. **

**6. In the figure, AC = AE, AB = AD and ∠ BAD = ∠ EAC. Show that BC = DE.**

**Sol. **∠BAD = ∠EAC [Given]

⇒ ∠BAD + ∠DAC = ∠EAC + ∠DAC [Adding ∠DAC to both sides]

⇒ ∠BAC = ∠DAE … (i)

Now, in ΔABC and ΔADE, we have

AB = AD [Given]

AC = AE [Given]

⇒ ∠BAC = ∠DAE [From (i)]

∴ ΔABC ≅ ΔADE [By SAS congruence]

⇒ BC = DE. [CPCT] Proved.

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

Show that

(i) ΔDAP ≅ ΔEBP (ii) AD = BE

**Sol. **In ΔDAP and ΔEBP, we have

AP = BP [Since P is the mid-point of line segment AB]

∠PAD = ∠PBE [Given]

∠EPB = ∠DPA

[since ∠EPA = ∠DPB ⇒ ∠EPA + ∠DPE = ∠DPB + ∠DPE]

∴ Δ DPA ≅ Δ EPB [By ASA congruence]

⇒ AD = BE [CPCT] **Proved. **