Chapter 7: Triangles (Exercise 7.1)
Chapter 7: Triangles (Exercise 7.1)
Question 1:
In quadrilateral ACBD, AC = AD and AB bisects ∠A. Show that ΔABC ≅ ΔABD. What can you say about BC and BD?
Solution:
* In ΔABC and ΔABD:
1. AC = AD (Given)
2. ∠CAB = ∠DAB (Since AB bisects ∠A)
3. AB = AB (Common side)
* By SAS Congruence Criterion:
ΔABC ≅ ΔABD
* Consequently, by CPCT (Corresponding Parts of Congruent Triangles):
BC = BD (BC and BD are of equal length).
Question 2:
ABCD is a quadrilateral in which AD = BC and ∠DAB = ∠CBA. Prove that:
1. ΔABD ≅ ΔBAC
2. BD = AC
3. ∠ABD = ∠BAC
Solution:
1. To prove ΔABD ≅ ΔBAC:
In ΔABD and ΔBAC:
* AD = BC (Given)
* ∠DAB = ∠CBA (Given)
* AB = BA (Common side)
Therefore, by SAS Congruence Rule, ΔABD ≅ ΔBAC.
2. To prove BD = AC:
Since ΔABD ≅ ΔBAC, by CPCT, BD = AC.
3. To prove ∠ABD = ∠BAC:
Since ΔABD ≅ ΔBAC, by CPCT, ∠ABD = ∠BAC.
(Hence proved.)
Question 3:
AD and BC are equal perpendiculars to a line segment AB. Show that CD bisects AB.
Solution:
* In ΔAOD and ΔBOC:
1. ∠OAD = ∠OBC = 90° (Given that AD and BC are perpendiculars)
2. ∠AOD = ∠BOC (Vertically opposite angles)
3. AD = BC (Given equal perpendiculars)
* By AAS Congruence Rule:
ΔAOD ≅ ΔBOC
* By CPCT:
AO = OB
* Since O is the midpoint of AB, line segment CD bisects AB.
(Hence proved.)
Question 4:
l and m are two parallel lines intersected by another pair of parallel lines p and q. Show that ΔABC ≅ ΔCDA.
Solution:
* Given l ∥ m and p ∥ q.
* In ΔABC and ΔCDA:
1. ∠BAC = ∠DCA (Alternate interior angles as l ∥ m with transversal AC)
2. AC = CA (Common side)
3. ∠BCA = ∠DAC (Alternate interior angles as p ∥ q with transversal AC)
* By ASA Congruence Rule:
ΔABC ≅ ΔCDA
(Hence proved.)
Chapter 7: Triangles (Exercise 7.1 - Remaining Questions)
Question 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:
1. ΔAPB ≅ ΔAQB
2. BP = BQ or B is equidistant from the arms of ∠A.
Solution:
1. To prove ΔAPB ≅ ΔAQB:
In ΔAPB and ΔAQB:
* ∠APB = ∠AQB = 90° (Since BP ⊥ AP and BQ ⊥ AQ)
* ∠PAB = ∠QAB (Since line l bisects ∠A)
* AB = AB (Common hypotenuse)
By AAS Congruence Rule, ΔAPB ≅ ΔAQB.
2. To prove BP = BQ:
Since ΔAPB ≅ ΔAQB, by CPCT:
BP = BQ (Hence, point B is equidistant from the arms of ∠A).
Question 6:
In the given figure, AC = AE, AB = AD and ∠BAD = ∠EAC. Show that BC = DE.
Solution:
* Given: ∠BAD = ∠EAC
* Adding ∠DAC to both sides:
∠BAD + ∠DAC = ∠EAC + ∠DAC
∠BAC = ∠DAE --- (Equation 1)
* Now, in ΔABC and ΔADE:
1. AB = AD (Given)
2. ∠BAC = ∠DAE (From Equation 1)
3. AC = AE (Given)
* By SAS Congruence Rule:
ΔABC ≅ ΔADE
* By CPCT:
BC = DE
(Hence proved.)
Question 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:
1. ΔDAP ≅ ΔEBP
2. AD = BE
Solution:
* Given that P is the midpoint of AB ⇒ AP = PB.
* Given: ∠EPA = ∠DPB
* Adding ∠DPE to both sides:
∠EPA + ∠DPE = ∠DPB + ∠DPE
∠DPA = ∠EPB --- (Equation 1)
1. To prove ΔDAP ≅ ΔEBP:
In ΔDAP and ΔEBP:
* ∠DAP = ∠EBP (Given ∠BAD = ∠ABE)
* AP = PB (Since P is the midpoint of AB)
* ∠DPA = ∠EPB (From Equation 1)
By ASA Congruence Rule, ΔDAP ≅ ΔEBP.
2. To prove AD = BE:
Since ΔDAP ≅ ΔEBP, by CPCT:
AD = BE
(Hence proved.)
Question 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:
1. ΔAMC ≅ ΔBMD
2. ∠DBC is a right angle
3. ΔDBC ≅ ΔACB
4. CM = ½ AB
Solution:
1. In ΔAMC and ΔBMD:
* AM = BM (Since M is the midpoint of AB)
* ∠AMC = ∠BMD (Vertically opposite angles)
* CM = DM (Given)
By SAS Congruence Rule, ΔAMC ≅ ΔBMD.
2. To prove ∠DBC = 90°:
Since ΔAMC ≅ ΔBMD, by CPCT, ∠ACM = ∠BDM.
Since these alternate interior angles are equal, line DB ∥ AC.
Therefore, interior angles on the same side sum to 180°:
∠ACB + ∠DBC = 180° ⇒ 90° + ∠DBC = 180° ⇒ ∠DBC = 90°.
3. In ΔDBC and ΔACB:
* DB = AC (From CPCT of ΔAMC ≅ ΔBMD)
* ∠DBC = ∠ACB = 90°
* BC = CB (Common side)
By SAS Congruence Rule, ΔDBC ≅ ΔACB.
4. To prove CM = ½ AB:
Since ΔDBC ≅ ΔACB, by CPCT, DC = AB.
Since CM = ½ DC (as M is midpoint of DC):
CM = ½ AB
(Hence proved.)