Chapter 7 Triangles Ex 7.1 Solutions
Question - 1 : - In quadrilateral ACBD, AC = AD and AB bisects ∠ A (see figure). Show that ∆ABC ≅ ∆ABD. What can you say about BC and BD?
Answer - 1 : -
In quadrilateral ACBD, we have AC = AD and AB being the bisector of ∠A.
Now, In ∆ABC and ∆ABD,
AC = AD (Given)
∠ CAB = ∠ DAB ( AB bisects ∠ CAB)
and AB = AB (Common)
∴ ∆ ABC ≅ ∆ABD (By SAS congruence axiom)
∴ BC = BD (By CPCT)
Question - 2 : - ABCD is a quadrilateral in which AD = BC and ∠ DAB = ∠ CBA (see figure). Prove that
(i) ∆ABD ≅ ∆BAC
(ii) BD = AC
(iii) ∠ABD = ∠ BAC
Answer - 2 : -
In quadrilateral ACBD, we have AD = BC and ∠ DAB = ∠ CBA
(i) In ∆ ABC and ∆ BAC,
AD = BC (Given)
∠DAB = ∠CBA (Given)
AB = AB (Common)
∴ ∆ ABD ≅ ∆BAC (By SAS congruence)
(ii) Since ∆ABD ≅ ∆BAC
⇒ BD =AC [By C.P.C.T.]
(iii) Since ∆ABD ≅ ∆BAC
⇒ ∠ABD = ∠BAC [By C.P.C.T.]
Question - 3 : - AD and BC are equal perpendiculars to a line segment AB (see figure). Show that CD bisects AB.
Answer - 3 : -
In ∆BOC and ∆AOD, we have
∠BOC = ∠AOD
BC = AD [Given]
∠BOC = ∠AOD [Vertically opposite angles]
∴ ∆OBC ≅ ∆OAD [By AAS congruency]
⇒ OB =OA [By C.P.C.T.]
i.e., O is the mid-point of AB.
Thus, CD bisects AB.
Question - 4 : - l and m are two parallellines intersected by another pair of parallel lines p and q (see figure). Showthat ∆ABC = ∆CDA.
Answer - 4 : -
∵ p || qand AC is a transversal,
∴ ∠BAC = ∠DCA …(1) [Alternate interior angles]
Also l || m and AC is a transversal,
∴ ∠BCA = ∠DAC …(2)
[Alternate interior angles]
Now, in ∆ABC and ∆CDA, we have
∠BAC = ∠DCA [From (1)]
CA = AC [Common]
∠BCA = ∠DAC [From (2)]
∴ ∆ABC ≅ ∆CDA [By ASA congruency]
Question - 5 : - Line l is the bisector of an ∠ A and ∠ B is any point on l. BP and BQ are perpendiculars from B to the arms ofLA (see figure). Show that
(i) ∆APB ≅ ∆AQB
(ii) BP = BQ or B is equidistant from the arms ot ∠A.
Answer - 5 : -
We have, l is the bisector of ∠QAP.
∴ ∠QAB = ∠PAB
∠Q = ∠P [Each 90°]
∠ABQ = ∠ABP
[By angle sum property of A]
Now, in ∆APB and ∆AQB, we have
∠ABP = ∠ABQ [Proved above]
AB = BA [Common]
∠PAB = ∠QAB [Given]
∴ ∆APB ≅ ∆AQB [By ASA congruency]
Since ∆APB ≅ ∆AQB
⇒ BP =BQ [By C.P.C.T.]
i. e., [Perpendicular distance of B from AP]
= [Perpendicular distance of B from AQ]
Thus, the point B is equidistant from the arms of ∠A.
Question - 6 : - In figure, AC = AE, AB = AD and ∠BAD = ∠EAC. Show that BC = DE.
Answer - 6 : -
We have, ∠BAD = ∠EAC
Adding ∠DAC onboth sides, we have
∠BAD + ∠DAC = ∠EAC + ∠DAC
⇒ ∠BAC = ∠DAE
Now, in ∆ABC and ∆ADE. we have
∠BAC = ∠DAE [Proved above]
AB = AD [Given]
AC = AE [Given]
∴ ∆ABC ≅ ∆ADE [By SAS congruency]
⇒ BC =DE [By C.P.C.T.]
Question - 7 : - AS is a line segment and P is its mid-point. D and Eare points on the same side of AB such that ∠BAD = ∠ ABE and ∠ EPA = ∠ DPB. (see figure). Showthat
(i) ∆DAP ≅∆EBP
(ii) AD = BE
Answer - 7 : -
We have, P is the mid-point of AB.
∴ AP = BP
∠EPA = ∠DPB [Given]
Adding ∠EPD onboth sides, we get
∠EPA + ∠EPD = ∠DPB + ∠EPD
⇒ ∠APD = ∠BPE
(i) Now, in ∆DAP and ∆EBP, we have
∠PAD = ∠PBE [ ∵∠BAD = ∠ABE]
AP = BP [Proved above]
∠DPA = ∠EPB [Proved above]
∴ ∆DAP ≅ ∆EBP [By ASA congruency]
(ii) Since, ∆ DAP ≅ ∆ EBP
⇒ AD =BE [By C.P.C.T.]
Question - 8 : - In right triangle ABC, right angled at C, M isthe mid-point of hypotenuse AB. C is joined to M and produced to a point D suchthat DM = CM. Point D is joined to point B (see figure). Show that
(i) ∆AMC ≅ ∆BMD
(ii) ∠DBC is a right angle
(iii) ∆DBC ≅ ∆ACB
Answer - 8 : - Since M is the mid – point of AB.
∴ BM = AM
(i) In ∆AMC and ∆BMD, we have
CM = DM [Given]
(ii) Since ∆AMC ≅ ∆BMD
⇒ ∠MAC = ∠MBD [By C.P.C.T.]
But they form a pair of alternate interior angles.
∴ AC ||DB
Now, BC is a transversal which intersects parallel lines AC and DB,
∴ ∠BCA + ∠DBC = 180° [Co-interior angles]
But ∠BCA =90° [∆ABC is right angled at C]
∴ 90° + ∠DBC = 180°
⇒ ∠DBC = 90°
(iii) Again, ∆AMC ≅ ∆BMD [Proved above]
∴ AC =BD [By C.P.C.T.]
Now, in ∆DBC and ∆ACB, we have
BD = CA [Proved above]
∠DBC = ∠ACB [Each 90°]
BC = CB [Common]
∴ ∆DBC ≅ ∆ACB [By SAS congruency]
(iv) As∆DBC ≅ ∆ACB
DC = AB [By C.P.C.T.]
But DM = CM [Given]
∴ CM = 
DC= AB
⇒ CM = AB