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.