Exercise 7.1 Solutions (Class 9)
Triangles
1. In quadrilateral ACBD, AC = AD and AB bisects ∠A. Show that ΔABC ≅ ΔABD. What can you say about BC and BD?
>Proof:
In ΔABC and ΔABD:
• AC = AD (Given)
• ∠CAB = ∠DAB (AB bisects ∠A)
• AB = AB (Common side)
By SAS (Side-Angle-Side) congruence rule, ΔABC ≅ ΔABD.
Since the triangles are congruent, their corresponding parts are equal.
Therefore, by CPCT (Corresponding Parts of Congruent Triangles), BC = BD.
Conclusion: BC and BD are equal in length.
2. ABCD is a quadrilateral in which AD = BC and ∠DAB = ∠CBA. Prove that (i) ΔABD ≅ ΔBAC (ii) BD = AC (iii) ∠ABD = ∠BAC.
>(i) ΔABD ≅ ΔBAC:
In ΔABD and ΔBAC:
• AD = BC (Given)
• ∠DAB = ∠CBA (Given)
• AB = BA (Common side)
By SAS congruence rule, ΔABD ≅ ΔBAC. (Proved)
(ii) BD = AC:
Since ΔABD ≅ ΔBAC, their corresponding parts are equal.
Therefore, BD = AC (by CPCT). (Proved)
(iii) ∠ABD = ∠BAC:
Since ΔABD ≅ ΔBAC, their corresponding angles are equal.
Therefore, ∠ABD = ∠BAC (by CPCT). (Proved)
3. AD and BC are equal perpendiculars to a line segment AB. Show that CD bisects AB.
>To Prove: CD bisects AB, which means we need to prove AO = BO.
Proof:
Consider ΔBOC and ΔAOD.
• ∠BOC = ∠AOD (Vertically opposite angles)
• ∠CBO = ∠DAO = 90° (Given they are perpendiculars)
• BC = AD (Given equal perpendiculars)
By AAS (Angle-Angle-Side) congruence rule, ΔBOC ≅ ΔAOD.
Since the triangles are congruent, their corresponding parts are equal.
Therefore, BO = AO (by CPCT). This means O is the midpoint of AB, so CD bisects AB.
(Proved)
4. l and m are two parallel lines intersected by another pair of parallel lines p and q. Show that ΔABC ≅ ΔCDA.
>Proof:
Since l || m and AC is a transversal, ∠BCA = ∠DAC (Alternate interior angles). —(1)
Since p || q and AC is a transversal, ∠BAC = ∠DCA (Alternate interior angles). —(2)
Now, in ΔABC and ΔCDA:
• ∠BCA = ∠DAC (From 1)
• AC = CA (Common side)
• ∠BAC = ∠DCA (From 2)
By ASA (Angle-Side-Angle) congruence rule, ΔABC ≅ ΔCDA.
(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. Show that: (i) ΔAPB ≅ ΔAQB (ii) BP = BQ or B is equidistant from the arms of ∠A.
>(i) ΔAPB ≅ ΔAQB:
In ΔAPB and ΔAQB:
• ∠BAP = ∠BAQ (Line l bisects ∠A)
• ∠BPA = ∠BQA = 90° (Given BP and BQ are perpendiculars)
• AB = AB (Common side)
By AAS congruence rule, ΔAPB ≅ ΔAQB. (Proved)
(ii) BP = BQ:
Since ΔAPB ≅ ΔAQB, their corresponding parts are equal.
Therefore, BP = BQ (by CPCT). This means point B is equidistant from the arms of ∠A. (Proved)
6. In Fig. 7.21, AC = AE, AB = AD and ∠BAD = ∠EAC. Show that BC = DE.
>Given: ∠BAD = ∠EAC.
Add ∠DAC to both sides:
∠BAD + ∠DAC = ∠EAC + ∠DAC
This implies ∠BAC = ∠EAD.
Now, consider ΔABC and ΔADE.
• AB = AD (Given)
• ∠BAC = ∠EAD (Proved above)
• AC = AE (Given)
By SAS congruence rule, ΔABC ≅ ΔADE.
Since the triangles are congruent, their corresponding parts are equal.
Therefore, BC = DE (by 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. Show that (i) ΔDAP ≅ ΔEBP (ii) AD = BE.
>(i) ΔDAP ≅ ΔEBP:
Given ∠EPA = ∠DPB. Add ∠EPD to both sides.
∠EPA + ∠EPD = ∠DPB + ∠EPD => ∠APD = ∠BPE.
Now, in ΔDAP and ΔEBP:
• ∠PAD = ∠PBE (Given as ∠BAD = ∠ABE)
• AP = BP (P is the mid-point of AB)
• ∠APD = ∠BPE (Proved above)
By ASA congruence rule, ΔDAP ≅ ΔEBP. (Proved)
(ii) AD = BE:
Since ΔDAP ≅ ΔEBP, their corresponding parts are equal.
Therefore, AD = BE (by CPCT). (Proved)
8. 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. Point D is joined to point B. Show that…
>(i) ΔAMC ≅ ΔBMD:
In ΔAMC and ΔBMD:
• AM = BM (M is mid-point of AB)
• ∠AMC = ∠BMD (Vertically opposite angles)
• CM = DM (Given)
By SAS congruence rule, ΔAMC ≅ ΔBMD. (Proved)
(ii) ∠DBC is a right angle:
Since ΔAMC ≅ ΔBMD, ∠ACM = ∠BDM (by CPCT).
These are alternate interior angles for lines AC and DB. Thus, DB || AC.
Now, since DB || AC and BC is a transversal, the sum of co-interior angles is 180°.
∠DBC + ∠ACB = 180° => ∠DBC + 90° = 180° => ∠DBC = 90°. (Proved)
(iii) ΔDBC ≅ ΔACB:
In ΔDBC and ΔACB:
• DB = AC (from part (i), CPCT)
• ∠DBC = ∠ACB = 90° (Proved)
• BC = CB (Common side)
By SAS congruence rule, ΔDBC ≅ ΔACB. (Proved)
(iv) CM = 1/2 AB:
Since ΔDBC ≅ ΔACB, their corresponding parts are equal. So, DC = AB.
We know CM = 1/2 DC (as DM=CM).
Substituting DC with AB, we get CM = 1/2 AB. (Proved)
