1In quadrilateral ACBD, AC = AD and AB bisects ∠A (see Fig. 7.16). Show that △ABC ≅ △ABD. What can you say about BC and BD?
Answer & Explanation:
In △ABC and △ABD:
AC = AD (Given)
∠CAB = ∠DAB (AB bisects ∠A)
AB = AB (Common)
∴ △ABC ≅ △ABD (SAS rule).
Hence, BC = BD (CPCT). Reference: Section 7.3, SAS Congruence Rule (Axiom 7.1).
Fig. 7.16: Quadrilateral ACBD with AB bisecting ∠A
2ABCD is a quadrilateral in which AD = BC and ∠DAB = ∠CBA (see Fig. 7.17). Prove that (i) △ABD ≅ △BAC (ii) BD = AC (iii) ∠ABD = ∠BAC.
Answer & Explanation:
In △ABD and △BAC:
AD = BC (Given)
∠DAB = ∠CBA (Given)
AB = BA (Common)
∴ △ABD ≅ △BAC (SAS rule).
⇒ BD = AC and ∠ABD = ∠BAC (CPCT). Reference: Section 7.3, SAS Rule and CPCT.
Fig. 7.17: Quadrilateral ABCD
3AD and BC are equal perpendiculars to a line segment AB (see Fig. 7.18). Show that CD bisects AB.
Answer & Explanation:
In △AOD and △BOC (O = intersection of CD and AB):
∠AOD = ∠BOC (Vertically opposite)
∠OAD = ∠OBC = 90° (Perpendiculars)
AD = BC (Given)
∴ △AOD ≅ △BOC (AAS rule).
⇒ AO = BO ⇒ CD bisects AB. Reference: Section 7.3, AAS Congruence Rule.
4l and m are two parallel lines intersected by another pair of parallel lines p and q (see Fig. 7.19). Show that △ABC ≅ △CDA.
Answer & Explanation:
Since l ∥ m and p ∥ q,
∠BAC = ∠DCA (Alternate angles)
∠BCA = ∠DAC (Alternate angles)
AC = CA (Common)
∴ △ABC ≅ △CDA (ASA rule). Reference: Section 7.3, ASA Congruence Rule (Theorem 7.1).
5Line 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) △APB ≅ △AQB (ii) BP = BQ.
Answer & Explanation:
In △APB and △AQB:
∠PAB = ∠QAB (l bisects ∠A)
∠APB = ∠AQB = 90° (Perpendiculars)
AB = AB (Common)
∴ △APB ≅ △AQB (AAS rule).
⇒ BP = BQ (CPCT). Reference: Section 7.3, AAS Rule; also converse of Example in Exercise 7.1 Q5.
6In Fig. 7.21, AC = AE, AB = AD and ∠BAD = ∠EAC. Show that BC = DE.
Answer & Explanation:
Given ∠BAD = ∠EAC.
Adding ∠DAC to both sides: ∠BAC = ∠DAE.
In △BAC and △DAE:
AB = AD, AC = AE, ∠BAC = ∠DAE.
∴ △BAC ≅ △DAE (SAS).
⇒ BC = DE (CPCT). Reference: Section 7.3, SAS Rule.
Fig. 7.21: Points D and E on extensions
7AB 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. 7.22). Show that (i) △DAP ≅ △EBP (ii) AD = BE.
Answer & Explanation:
Given ∠EPA = ∠DPB.
Adding ∠EPD: ∠APD = ∠BPE.
In △DAP and △EBP:
∠DAP = ∠EBP (Given)
AP = BP (P mid-point)
∠APD = ∠BPE (Proved)
∴ △DAP ≅ △EBP (ASA).
⇒ AD = BE (CPCT). Reference: Section 7.3, ASA Rule.
8In right triangle ABC, right angled at C, M is the mid-point of hypotenuse AB. C is joined to M and produced to D such that DM = CM. Show that: (i) △AMC ≅ △BMD (ii) ∠DBC is a right angle (iii) △DBC ≅ △ACB (iv) CM = ½ AB.
Answer & Explanation:
(i) In △AMC and △BMD:
AM = BM (M mid-point)
CM = DM (Given)
∠AMC = ∠BMD (Vertically opposite)
∴ △AMC ≅ △BMD (SAS).
(ii) From (i), ∠ACM = ∠BDM ⇒ AC ∥ BD.
Since ∠ACB = 90°, ∠DBC = 90° (Co-interior angles).
(iii) In △DBC and △ACB:
BC = CB (Common), ∠DBC = ∠ACB = 90°, BD = AC (CPCT from (i))
∴ △DBC ≅ △ACB (SAS).
(iv) From (iii), DC = AB ⇒ 2CM = AB ⇒ CM = ½ AB. Reference: Section 7.3, SAS Rule and properties of parallel lines.
9In an isosceles triangle ABC, with AB = AC, the bisectors of ∠B and ∠C intersect at O. Join A to O. Show that: (i) OB = OC (ii) AO bisects ∠A.
Answer & Explanation:
(i) Since AB = AC, ∠B = ∠C (Theorem 7.2).
BO and CO bisect ∠B and ∠C ⇒ ∠OBC = ∠OCB.
∴ OB = OC (Theorem 7.3).
(ii) In △AOB and △AOC:
AB = AC, OB = OC, AO = AO ⇒ △AOB ≅ △AOC (SSS).
⇒ ∠BAO = ∠CAO ⇒ AO bisects ∠A. Reference: Section 7.4, Theorems 7.2 and 7.3.
10In △ABC, AD is the perpendicular bisector of BC (see Fig. 7.30). Show that △ABC is isosceles with AB = AC.
Answer & Explanation:
In △ABD and △ACD:
BD = CD (D mid-point), ∠ADB = ∠ADC = 90°, AD = AD.
∴ △ABD ≅ △ACD (SAS).
⇒ AB = AC. Reference: Section 7.4, Example 4 and SAS Rule.
11△ABC is isosceles with altitudes BE and CF to equal sides AC and AB. Show BE = CF.
Answer & Explanation:
In △ABE and △ACF:
∠A = ∠A (Common), ∠AEB = ∠AFC = 90°, AB = AC.
∴ △ABE ≅ △ACF (AAS).
⇒ BE = CF (CPCT). Reference: Section 7.4, AAS Rule.
12In △ABC, altitudes BE and CF to sides AC and AB are equal. Show that: (i) △ABE ≅ △ACF (ii) AB = AC.
Answer & Explanation:
(i) In △ABE and △ACF:
∠A = ∠A, ∠AEB = ∠AFC = 90°, BE = CF (Given).
∴ △ABE ≅ △ACF (AAS).
(ii) ⇒ AB = AC (CPCT). Reference: Section 7.4, AAS Rule and CPCT.
13△ABC and △DBC are two isosceles triangles on the same base BC (see Fig. 7.33). Show that ∠ABD = ∠ACD.
Answer & Explanation:
AB = AC and DB = DC.
In △ABD and △ACD:
AB = AC, BD = CD, AD = AD ⇒ △ABD ≅ △ACD (SSS).
⇒ ∠ABD = ∠ACD (CPCT). Reference: Section 7.4, SSS Rule and CPCT.
14△ABC is isosceles with AB = AC. BA is produced to D such that AD = AB. Show that ∠BCD is a right angle.
Answer & Explanation:
AB = AC = AD.
In △ABC, ∠ABC = ∠ACB = x.
In △ACD, ∠ACD = ∠ADC = y.
In △BCD, angles sum: x + y + (x + y) = 180° ⇒ 2(x + y) = 180° ⇒ x + y = 90°.
∴ ∠BCD = ∠ACB + ∠ACD = x + y = 90°. Reference: Section 7.4, Angle sum property and isosceles triangle properties.
15△ABC is a right angled triangle in which ∠A = 90° and AB = AC. Find ∠B and ∠C.
Answer & Explanation:
AB = AC ⇒ ∠B = ∠C.
∠A + ∠B + ∠C = 180° ⇒ 90° + 2∠B = 180° ⇒ ∠B = ∠C = 45°. Reference: Section 7.4, Theorem 7.2 and angle sum property.
16Show that the angles of an equilateral triangle are 60° each.
Answer & Explanation:
In equilateral triangle, AB = BC = CA ⇒ ∠A = ∠B = ∠C.
∠A + ∠B + ∠C = 180° ⇒ 3∠A = 180° ⇒ ∠A = 60°.
∴ Each angle is 60°. Reference: Section 7.6, Point 10.
17△ABC and △DBC are isosceles on same base BC, vertices A and D on same side. AD extended meets BC at P. Show that: (i) △ABD ≅ △ACD (ii) △ABP ≅ △ACP (iii) AP bisects ∠A and ∠D (iv) AP is perpendicular bisector of BC.
Answer & Explanation:
(i) AB = AC, DB = DC, AD = AD ⇒ △ABD ≅ △ACD (SSS).
(ii) From (i), ∠BAP = ∠CAP. In △ABP and △ACP: AB=AC, ∠BAP=∠CAP, AP=AP ⇒ △ABP≅△ACP (SAS).
(iii) ∠BAP = ∠CAP ⇒ AP bisects ∠A. Similarly, from (i), ∠ADP = ∠ACP ⇒ AP bisects ∠D.
(iv) From (ii), BP = CP and ∠APB = ∠APC = 90° (Linear pair). So AP ⊥ BC and bisects it. Reference: Section 7.5, SSS and SAS Rules.
18AD is altitude of isosceles △ABC (AB = AC). Show that: (i) AD bisects BC (ii) AD bisects ∠A.
Answer & Explanation:
In △ABD and △ACD:
AB = AC, ∠ADB = ∠ADC = 90°, AD = AD ⇒ △ABD ≅ △ACD (RHS).
⇒ BD = CD and ∠BAD = ∠CAD. Reference: Section 7.5, RHS Congruence Rule (Theorem 7.5).
19Two sides AB, BC and median AM of △ABC equal to PQ, QR and median PN of △PQR. Show: (i) △ABM ≅ △PQN (ii) △ABC ≅ △PQR.
Answer & Explanation:
(i) AB = PQ, BC = QR ⇒ BM = QN (medians to equal sides). AM = PN (Given).
∴ △ABM ≅ △PQN (SSS).
(ii) From (i), ∠B = ∠Q. In △ABC and △PQR: AB=PQ, BC=QR, ∠B=∠Q ⇒ △ABC ≅ △PQR (SAS). Reference: Section 7.5, SSS and SAS Rules.
20BE and CF are two equal altitudes of △ABC. Using RHS congruence rule, prove that △ABC is isosceles.
Answer & Explanation:
In △BEC and △CFB:
BE = CF (Given), BC = CB (Common), ∠BEC = ∠CFB = 90°.
∴ △BEC ≅ △CFB (RHS).
⇒ ∠B = ∠C ⇒ AB = AC (Theorem 7.3). Reference: Section 7.5, RHS Rule and Theorem 7.3.
21△ABC is isosceles with AB = AC. Draw AP ⊥ BC to show that ∠B = ∠C.
Answer & Explanation:
In △ABP and △ACP:
AB = AC, AP = AP, ∠APB = ∠APC = 90°.
∴ △ABP ≅ △ACP (RHS).
⇒ ∠B = ∠C (CPCT). Reference: Section 7.5, RHS Rule and CPCT.