In ΔOCB and ΔODA, ∠BOC = ∠AOD [∵ Vertically Opposite Angles] ∠CBO = ∠DAO [∵ Each 90°] BC = AD [ Given] Hance, ΔOCB ≅ ΔODA [∵ AAS Congruency Rule] BO = AO [∵ Corresponding parts of congruent triangles are equal] Hence. CD bisects AB.
In ΔOCB and ΔODA,
∠BOC = ∠AOD [∵ Vertically Opposite Angles]
∠CBO = ∠DAO [∵ Each 90°]
BC = AD [ Given]
Hance, ΔOCB ≅ ΔODA [∵ AAS Congruency Rule]
BO = AO [∵ Corresponding parts of congruent triangles are equal]
Hence. CD bisects AB.
(i) In ΔAPB and ΔAQB, ∠APB = ∠AQB [∵ Each 90°] ∠PAB = ∠QAB [∵ Line l bisects angle A] AB = AB [∵ Common] Hance, ΔAPB ≅ ΔAQB [∵ ASA Congruency Rule] (ii) BP = BQ [∵ Corresponding parts of congruent Triangles are equal ]
(i) In ΔAPB and ΔAQB,
∠APB = ∠AQB [∵ Each 90°]
∠PAB = ∠QAB [∵ Line l bisects angle A]
AB = AB [∵ Common]
Hance, ΔAPB ≅ ΔAQB [∵ ASA Congruency Rule]
(ii) BP = BQ [∵ Corresponding parts of congruent Triangles are equal ]
∠BAD = ∠EAC [∵ Given] Adding ∠DAC both sides, we have ∠BAD + ∠DAC = ∠EAC + ∠DAC ⇒ ∠BAC = ∠EAD In ΔBAC and ΔDAE, AB = AD [∵ Given] ∠BAC = ∠EAD [∵ Proved above] AC = AE [∵ Given] Hance, ΔBAC ≅ ΔDAE [∵ SAS Congruency Rule] BC = DE [∵ Corresponding parts of congruent Triangles are equal ]
∠BAD = ∠EAC [∵ Given]
Adding ∠DAC both sides, we have
∠BAD + ∠DAC = ∠EAC + ∠DAC
⇒ ∠BAC = ∠EAD
In ΔBAC and ΔDAE,
AB = AD [∵ Given]
∠BAC = ∠EAD [∵ Proved above]
AC = AE [∵ Given]
Hance, ΔBAC ≅ ΔDAE [∵ SAS Congruency Rule]
BC = DE [∵ Corresponding parts of congruent Triangles are equal ]
(i) In ΔAMC and ΔBMD, CM = DN [∵ Given] ∠AMC = ∠BMD [∵ Vertically Opposite Angles] AM = MB [∵ M is the mid-point of line segment AB] Hance, ΔAMC ≅ ΔBMD [∵ SAS Congruency Rule] (ii) ΔAMC ≅ ΔBMD [∵ Proved above] ∠CAM = ∠DMB [∵ Corresponding parts of congruent Triangles are equal ] Since, alternate angRead more
(i) In ΔAMC and ΔBMD,
CM = DN [∵ Given]
∠AMC = ∠BMD [∵ Vertically Opposite Angles]
AM = MB [∵ M is the mid-point of line segment AB]
Hance, ΔAMC ≅ ΔBMD [∵ SAS Congruency Rule]
(ii) ΔAMC ≅ ΔBMD [∵ Proved above]
∠CAM = ∠DMB [∵ Corresponding parts of congruent Triangles are equal ]
Since, alternate angles (∠CAM and ∠DMB) are equal, therefore AC ∥ BD.
∠ACB + ∠DBC = 180° [∵ Co-interior angles]
⇒ 90° + DBC = 180° [∵ Angle C is right angle]
⇒ ∠DBC = 180° – 90° = 90°
Hence, ∠DBC is a right angle.
(iii) In ΔDBC and ΔACB
DB = AC [∵ ΔAMC ≅ ΔBMD]
∠DBC = ∠ACB [∵ Proved above]
BC = BC [∵ M is the mid-point of line segment AB]
Hence, ΔDBC ≅ ΔACB [∵ SAS Congruency Rule]
AD and BC are equal perpendiculars to a line segment AB (see Figure). Show that CD bisects AB.
In ΔOCB and ΔODA, ∠BOC = ∠AOD [∵ Vertically Opposite Angles] ∠CBO = ∠DAO [∵ Each 90°] BC = AD [ Given] Hance, ΔOCB ≅ ΔODA [∵ AAS Congruency Rule] BO = AO [∵ Corresponding parts of congruent triangles are equal] Hence. CD bisects AB.
In ΔOCB and ΔODA,
See less∠BOC = ∠AOD [∵ Vertically Opposite Angles]
∠CBO = ∠DAO [∵ Each 90°]
BC = AD [ Given]
Hance, ΔOCB ≅ ΔODA [∵ AAS Congruency Rule]
BO = AO [∵ Corresponding parts of congruent triangles are equal]
Hence. CD bisects AB.
l and m are two parallel lines intersected by another pair of parallel lines p and q (see Figure). Show that ΔABC ≅ ΔCDA.
In ΔABC and ΔCDA, ∠BAC = ∠ACD [∵ Alternate Angles] AC = AC [∵ Common] ∠BCA = ∠DAC [∵ Alternate Angles] Hance, ΔABC ≅ ΔCDA [∵ ASA Congruency Rule]
In ΔABC and ΔCDA,
See less∠BAC = ∠ACD [∵ Alternate Angles]
AC = AC [∵ Common]
∠BCA = ∠DAC [∵ Alternate Angles]
Hance, ΔABC ≅ ΔCDA [∵ ASA Congruency Rule]
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 Figure). Show that: (i) Δ APB ≅ Δ AQB (ii) BP = BQ or B is equidistant from the arms of ∠ A.
(i) In ΔAPB and ΔAQB, ∠APB = ∠AQB [∵ Each 90°] ∠PAB = ∠QAB [∵ Line l bisects angle A] AB = AB [∵ Common] Hance, ΔAPB ≅ ΔAQB [∵ ASA Congruency Rule] (ii) BP = BQ [∵ Corresponding parts of congruent Triangles are equal ]
(i) In ΔAPB and ΔAQB,
∠APB = ∠AQB [∵ Each 90°]
∠PAB = ∠QAB [∵ Line l bisects angle A]
AB = AB [∵ Common]
Hance, ΔAPB ≅ ΔAQB [∵ ASA Congruency Rule]
(ii) BP = BQ [∵ Corresponding parts of congruent Triangles are equal ]
See lessIn Figure, AC = AE, AB = AD and ∠ BAD = ∠ EAC. Show that BC = DE.
∠BAD = ∠EAC [∵ Given] Adding ∠DAC both sides, we have ∠BAD + ∠DAC = ∠EAC + ∠DAC ⇒ ∠BAC = ∠EAD In ΔBAC and ΔDAE, AB = AD [∵ Given] ∠BAC = ∠EAD [∵ Proved above] AC = AE [∵ Given] Hance, ΔBAC ≅ ΔDAE [∵ SAS Congruency Rule] BC = DE [∵ Corresponding parts of congruent Triangles are equal ]
∠BAD = ∠EAC [∵ Given]
See lessAdding ∠DAC both sides, we have
∠BAD + ∠DAC = ∠EAC + ∠DAC
⇒ ∠BAC = ∠EAD
In ΔBAC and ΔDAE,
AB = AD [∵ Given]
∠BAC = ∠EAD [∵ Proved above]
AC = AE [∵ Given]
Hance, ΔBAC ≅ ΔDAE [∵ SAS Congruency Rule]
BC = DE [∵ Corresponding parts of congruent Triangles are equal ]
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 (see Figure). Show that: (i) Δ AMC ≅ Δ BMD (ii) ∠ DBC is a right angle. (iii) Δ DBC ≅ Δ ACB (iv) CM = (1/2) AB
(i) In ΔAMC and ΔBMD, CM = DN [∵ Given] ∠AMC = ∠BMD [∵ Vertically Opposite Angles] AM = MB [∵ M is the mid-point of line segment AB] Hance, ΔAMC ≅ ΔBMD [∵ SAS Congruency Rule] (ii) ΔAMC ≅ ΔBMD [∵ Proved above] ∠CAM = ∠DMB [∵ Corresponding parts of congruent Triangles are equal ] Since, alternate angRead more
(i) In ΔAMC and ΔBMD,
CM = DN [∵ Given]
∠AMC = ∠BMD [∵ Vertically Opposite Angles]
AM = MB [∵ M is the mid-point of line segment AB]
Hance, ΔAMC ≅ ΔBMD [∵ SAS Congruency Rule]
(ii) ΔAMC ≅ ΔBMD [∵ Proved above]
∠CAM = ∠DMB [∵ Corresponding parts of congruent Triangles are equal ]
Since, alternate angles (∠CAM and ∠DMB) are equal, therefore AC ∥ BD.
∠ACB + ∠DBC = 180° [∵ Co-interior angles]
⇒ 90° + DBC = 180° [∵ Angle C is right angle]
⇒ ∠DBC = 180° – 90° = 90°
Hence, ∠DBC is a right angle.
(iii) In ΔDBC and ΔACB
See lessDB = AC [∵ ΔAMC ≅ ΔBMD]
∠DBC = ∠ACB [∵ Proved above]
BC = BC [∵ M is the mid-point of line segment AB]
Hence, ΔDBC ≅ ΔACB [∵ SAS Congruency Rule]