Triangles-NCERT Solutions

1.  In quadrilateral ACBD, AC = AD and AB bisects ∠A )see figure). Show that ΔABC ≌ ΔABD. What can you say about BC and BD?

Ans. In quadrilateral ABCD we have

                  AC = AD

            and AB being the bisector of ∠A.

            Now, in ΔABC and ΔABD,

                  AC = AD

[Given]

                  AB = AB

[Common]

∠CAB = ∠DAB [∴ AB bisects ∠CAD]

            ∴ Using SAS criteria, we have

                  ΔABC ≌ ΔABD.

            ∴ Corresponding parts of congruent triangles (c.p.c.t) are equal.

            ∴ BC = BD.

2.  ABCD is a quadrilateral in which AD = BC and ∠DAB = ∠CBA (see Figure). Prove that

      (i) ΔABD ≌ ΔBAC

      (ii) BD = AC

      (iii) ∠ABD = ∠BAC.

Ans. (i) In quadrilateral ABCD, we have AD = BC and

                ∠DAB = ∠CBA.

                In ΔABD and ΔBAC,

                AD = BC

[Given]

                AB = BA

[Common]

                ∠DAB = ∠CBA

[Given]

                ∴ Using SAS criteria, we have ΔABD ≌ ΔBAC

          (ii) ∵ ΔABD ≌ ΔBAC

                ∴ Their corresponding parts are equal.

                ⇒ BD = AC

          (ii) Since ΔABD ≌ ΔBAC

                ∴ Their corresponding parts are equal.

                ⇒ ∠ABD = ∠BAC.

3.  AD and BC are equal perpendiculars to a line segment AB (see figure). Show that CD bisects AB.

Ans. We have ∠ABC = 90° and ∠BAD = 90°

          Also AB and CD intersect at O.

          ∴ Vertically opposite angles are equal.

          Now, in ΔOBC and ΔOAD, we have

                  ∠ABC = ∠BAD

[each = 90°]

                  BC = AD

[Given]

                  ∠BOC = ∠AOD

[vertically opposite angles]

          ∴ Using ASA criteria, we have

                  ΔOBC ≌ ΔOAD

          ⇒ OB = OA

[c.p.c.t]

          i.e. O is the mid-point of AB

          Thus, CD bisects AB.

4.  l and m are two parallel lines intersected by another pair of parallel lines p and q (see figure). Show that ΔABC ≌ ΔCDA.

Ans. ∵l || m and AC is a transversal.

          ∴ ∠BAC = ∠DCA

[Alternate interior angles]

          Also p || q and AC is a transversal,

          ∴ ∠BCA = ∠DAC

[Alternate interior angles]

          Now, in ΔABC and ΔCDA,

                ∠BAC = ∠DCA

[Proved]

                ∠BCA = ∠DAC

[Proved]

                CA = AC

[Common]

          ∴ Using ASA criteria, we have ΔABC ≌ ΔCDA

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 (see Figure). Show that:

          (i) ΔAPB ≌ ΔAQB

          (ii) BO = BQ or B is equidistant from the arms of ∠A.

Ans. We have, l as the bisector of QAP.

          ∴ ∠QAB = ∠PAB

          ∴ ∠Q = ∠P

[each = 90°]

          ⇒ Third ∠ABQ = Third ∠ABP

          (i) Now, in ΔAPB and ΔAQB, we have

                AB = AB

[Common]

                ∠ABP = ∠ABQ

[Proved]

                ∠PAB = ∠QAB

[Proved]

                ∴ Using SAS criteria, we have

                ΔAPB ≌ ΔAQB

          (ii) Since ΔAPB ≌ ΔAQB

                ∴ Their corresponding angles are equal.

                ⇒ BP = BQ

                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.

6.  In the figure, AC = AE, AB = AD and ∠BAD = ∠ESC. Show that BC = DE.

Ans. We have ∠BAD = ∠EAC

            Adding ∠DAC on both sides, we have

            ∠BAD + ∠DAC = ∠EAC + ∠DAC

            ⇒ ∠BAC = ∠DAE

            Now, in ΔABC and ΔADE, we have

                ∠BAC = ∠DAE

[Proved]

                AB = AD

[Given]

                AC = AE

[Given]

            ∴ ΔABC ≌ ΔADE

[Using SAS criteria]

            Since ΔABC ≌ ΔADE, therefore, their corresponding parts are equal.

            ⇒ BC = DE.

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 (see figure). Show that

      (i) ΔDAP ≌ ΔEBP

      (ii) AD = BE

Ans. We have, P is the mid-point of AB.

            AP = BP

            ∠EPA = ∠DPB

[Given]

            Adding ∠EPD on both sides, we get

            ∠EPA + ∠EPD = ∠DPB + ∠EPD

            ⇒ APD = ∠BPE

            (i) Now, in ΔDAP ≌ ΔEBP, we have

                  AP = BP

[Proved]

            ∠PAD = ∠ PBE

[∴ It is given that ∠BAD = ∠ABE]

            ∠DPA = ∠EPB

[Proved]

            ∴ Using ASA criteria, we have

            ΔDAP ≌ ΔEBP

            (ii) Since,

                 ∴ Their corresponding parts are equal.

                 ⇒ AD = BE.

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 (see figure). Show that

      (i) ΔAMC ≌ ΔBMD

      (ii) ∠DBC is a right angle.

      (ii) ΔDBC ≌ ΔACB

      (iv) CM = AB

Ans. ∵ M is the mid-point of AB.

            ∴ BM = AM

[Given]

            (i) In ΔAMC and ΔBMD, we have

                CM = DM

[Given]

                AM = BM

[Proved]

                ∠AMC = ∠BMD

[Vertically opposite angles]

                ∴ ΔAMC ≌ ΔBMD

[SAS criteria]

            (ii) ∴ΔAMC ≌ ΔBMD

                ∴ Their corresponding parts are equal.

                ⇒ ∠MAC = ∠MBD

                But they form a pair of alternate interior angles.

                ∴ AC || DB

                Now, BC is a transversal which intersecting parallel lines AC and DB,

                ∴ ∠BCA + ∠DBC = 180°

                But ∠BCA = 90°

[∴ ΔABC is right angled at C]

                ∴ 90° + ∠DBC = 180°

                or ∠DBC = 180° – 90° = 90°

                Thus, ∠DBC = 90°

            (iii) Again, ΔAMC ≌ ΔBMD

[Proved]

                ∴ AC = BD

[c.p.c.t]

                Now, in ΔDBC and ΔACB, we have

                ∠DBC = ∠ACB

[Each = 90°]

                BD = CA

[Proved]

                BC = CB

[Common]

                ∴ Using SAS criteria, we have

                            ΔDBC ≌ ΔACB

            (iv) ∴ ΔDBC ≌ ΔACB

                ∴ Their corresponding parts are equal.

                ⇒ DC = AB

                But DM = CM

[Given]

                

1.  In an isosceles triangle ABC, with AB = AC, the bisectors of ∠B and ∠C intersect each other at O. Join A to O. Show that:

      (i) OB = OC                  (ii) AO bisects ∠A

Ans. (i) In ΔABC, we have

                ∴ AB = AC

[Given]

                ∠C = ∠B

                [Angle opposite to equal sides are equal]

                

                or ∠OCB = ∠OBC

                ⇒ OB = OC

                [Sides opposite to equal angles are equal]

          (ii) In ΔABO and ΔACO, we have

                AB = AC

[Given]

                OB = OC

[Proved]

                

                ∴ Using SAS criteria,

                ΔABO ≌ ΔACO

                ⇒ ∠OAB = ∠OAC

[c.p.c.t.]

                ⇒ AO bisects ∠A.

2.  In ΔABC, AD is the perpendicular bisector of BC (see figure). Show that ΔABC is an isosceles triangle in which AB = AC.

Ans. ∵AD is bisector of BC.

           ∴ BD = CD

           Now, in ΔABD and ΔACD, we have:

           AD = AD

[Common]

           ∠ADB = ∠ADC = 90°

[∴ AD ⊥ BC]

           BD = CD

[Proved]

           ∴ ΔABD ≌ ΔACD

[SAS criteria]

           ∴ Their corresponding parts are equal.

           ⇒ AB = AC

           Thus, ΔABC is an isosceles triangle.

3.  ABC is an isosceles triangle in which altitudes BE and CF are drawn to equal sides AC and AB respectively (see Figure) Show that these altitudes are equal.

OR

ABC is an isosceles triangle with AB = AC. Prove that the altitudes BE and CF of the triangle are equal.

Ans. ΔABC is an isosceles triangle.

          ∴ AB = AC

          ⇒ ∠ACB = ∠ABC

[∴ Angles opposite to equal sides are equal]

          Now, in ΔBEC and ΔCFB, we have

          ∠EBC = ∠FCB

[Proved]

          BC = CB

[Common]

          and ∠BEC = ∠CFB

[Each = 90°]

          ∴ ΔBEC ≌ ΔCFB

[Using ASA criteria]

          ⇒ Their corresponding parts are equal.

          i.e. BE = CF

4.  ABC is a triangle is which altitudes BE and CF to sides AC and AB are equal (see figure). Show that

      (i) ΔABE ≌ ΔACF

      (ii) AB = AC, i.e. ABC is an isosceles triangle.

Ans. (i) In ΔABE and ΔACF, we have

          ∠AEB = ∠AFC

          [each = 90° ∵ BE ⊥AC and CF ⊥ AB]

          ∠A = ∠A

[Common]

          BE = CF

[Given]

          ∴ ΔABE ≌ ΔACF

[Using AAS criterion]

          (ii) Since, ABE ≌ ΔACF

                  ∴ Their corresponding parts are equal.

                  ⇒ AB = AC

5.  ABC and DBC are two isosceles triangles on the same base BC (see figure). Show that ∠ABD = ∠ACD.

Ans. In ΔABC, we have

AB = AC [∴ ΔABC is an isosceles triangle]

          But angles opposite to equal sides are equal.

          ∴ ∠ABC = ∠ACB

...(1)

          Again, in ΔBDC, we have

BD = CD [∴ ΔBDC is an isosceles triangle.]

          ∴ ∠CBD = ∠BCD

...(2)

[Angles opposite to equal sides are equal]

          Adding (1) and (2), we have

          ∠ABC + ∠CBD = ∠ACB + ∠BCD

          ⇒ ∠ABD = ∠ACD

6.  ΔABC is an isosceles triangle in which AB = AC. Side BA is produced to D such that AD = AB (see figure). Show that ∠BCD is a right angle.

Ans. In ΔABC,

          From (1) and (2), we have

          AC = AD

          Now, in ΔABC, we have

          ∠B + ∠ACB + ∠BAC = 180°

          ⇒ 2∠ACB + ∠BAC = 180°

...(3)

[∴ ∠B = ∠ACB (Angles opposite to equal sides)]

          In ΔACD,

          ∠D + ∠ACD + ∠CAD = 180°

          ⇒ 2∠ACD + ∠CAD = 180°

[∴ ∠D = ∠ACD (angles opposite to equal sides)]

          Adding (3) and (4), we have

          2∠ACB + ∠BAC + 2∠ACD + ∠CAD = 180° + 180°

          ⇒ 2[∠ACB + ∠ACD] + [∠BAC + ∠CAD] = 360°

          ⇒ 2[∠BCD] + [180°] = 360° [∠BAC and ∠CAD form a linear pair]

          ⇒ 2∠BCD = 360° – 180° = 180°

          

          Thus, ∠BCD = 90°.

7.  ABC is a right angled triangle in which ∠A = 90° and AB = AC. Find ∠B and ∠C.

Ans. In ΔABC, we have

          AB = AC

[Given]

          ∴ Their opposite angles are equal.

          ⇒∠ACB = ∠ABC

          Now, ∠A + ∠B + ∠C = 180°

          ⇒90° + ∠B + ∠C = 180°

[∴ ∠A = 90° (Given)]

          ⇒ ∠B + ∠C = 180°

          But ∠ABC = ∠ACB, i.e. ∠B = ∠C

          

          Thus, ∠B = 45° and ∠C = 45°

8.  Show that the angles of an equilateral triangle of 60° each.

Ans. IN ΔABC, we have

AB = BC = CA [∴ ABC is an equilateral triangle]

          ∴ AB = BC ⇒ ∠A = ∠C

...(1)

[∴ Angle opposite to equal sides are equal.]

          Similarly, AC = BC ⇒ ∠A = ∠B

...(2)

          From (1) and (2), we have

          ∠A = ∠B = ∠C

          Let ∠A = ∠B = ∠C = x

          Since, ∠A + ∠B + ∠C = 180°

          ∴ x + x + x = 180°

          ⇒ 3x = 180°

          

          Thus, the angles of an equilateral triangle are 60° each.

1.  ΔABC and ΔDBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC (see figure). If AD is extended to intersect BC at P, show that

      (i) ΔABD ≌ ΔACD

      (ii) ΔABP ≌ ΔACP

      (iii) AP bisects ∠A as well as ∠D.

      (iv) AP is the perpendicular bisector of BC.

Ans. (i) In ΔABD and ΔACD, we have

                AB = AC

                AD = AD

                BD = CD

                ΔABD ≌ ΔACD

            (ii) In ΔABP and ΔACP, we have

                AB = AC

[Given]

                ∴ AB = AC ⇒ ∠B = ∠C

[∴ Angle opposite to equal sides are equal]

                AP = AP

[Common]

                ∴ ΔABP ≌ ACP

[SAS Criteria]

            (iii) Since, ΔABP ≌ ΔACP

                ∴ Their corresponding parts are congruent.

                ⇒ ∠BAP = ∠CAP

                ∴ AP is the bisector of ∠A.

...(1)

                Again, in ΔBDP and ΔCDP, we have

                BD = CD

[Given]

                ∠DBP = ∠CDP

[Angles opposite to equal sides]

                DP = DP

[Common]

                ⇒ ∠BDP ≌ ∠CDP

                ∴ ∠BDP ≌ ∠CDP [c.p.c.t]

                ⇒ DP (or AP) is the bisector of ∠D.

...(2)

                From (1) and (2), Ap is the bisector of ∠A as well as ∠D.

            (iv) ∵ΔABP ≌ ΔACP

                ∴ Their corresponding parts are equal.

                ⇒∠APB = ∠APC

But ∠APB + ∠APC = 180° [Linear pair]

                ∴ ∠APB = ∠APC = 90°

                ⇒ AP ⊥ BC

                ⇒ AP is the perpendicular bisector of BC

2.  AD is an altitude of an isosceles triangle ABC in which AB = AC. Show that

      (i) AD bisects BC                        (ii) AD bisects ∠A

Ans. (i) In ΔABD and ΔACD, we have

                AB = AC

[Given]

                ∠B = ∠C

[Angles opposite to equal sides]

                AD = AD

[Common]

                ∴ ΔABD ≌ ΔACD

                ⇒ Their corresponding parts are equal.

                ∴ BD = CD

                ⇒ D is the mid-point of BC or AD bisects BC.

          (ii) Since, ΔABD ≌ ΔACD,

                ∴ Their corresponding parts are congruent.

                ⇒ ∠BAD = ∠CAD

                ⇒ AD bisects ∠A.

3.  Two sides AB and BC and median AM of one triangle ABC are respectively equal to sides PQ and QR and median PN of ΔPQR (see figure). Show that

      (i) ΔABM ≌ ΔPQN                        (ii) ΔABC ≌ ΔPQR

Ans. In ΔABC, AM is a median

[Given]

            

...(1)

            In ΔPQR, PN is a median.

            

...(2)

            ∴ BC = QR

[Given]

            

            ⇒BM =QN

[From (1) and (2)]

          (i) In ΔABM and ΔPQN, we have

                ∴AB = PQ

[Given]

                AM = PN

[Given]

                BM = QN

[Proved]

                ∴ ΔABM ≌ ΔPQN

[SSS criteria]

          (ii) ∵ΔABM ≌ ΔPQN

                ∴ Their corresponding parts are congruent.

                ⇒ ∠B = ∠Q

                Now, in ΔABC and ΔPQR, we have

                ∠B = ∠Q

[Proved]

                AB = PQ

[Given]

                BC = QR

[Given]

                ΔABC ≌ ΔPQR

4.  BE and CF are two equal altitudes of a triangle ABC. Using RHS congruence rule, prove that the triangle ABC is isosceles.

Ans. BE ⊥ AC

[Given]

          ∴ ΔBEC is a right triangle such that

          ∠BEC = 90°

          Similarly, ∠CFB = 90°

          Now, in right ΔBEC and right ΔCFB, we have

          BE = CF

[Given]

          BC = CB

[Common]

          ∴ Using RHS criteria, ΔBEC ≌ ΔCFB

          ∴ Their corresponding parts are equal.

          ⇒ ∠BCE = ∠CBF

          or ∠BCA = ∠CBA

          Now, in ΔABC, ∠BCA = ∠CBA

          ∴ Their opposite sides are equal.

          ⇒ AB = AC

          ∴ ΔABC is an isosceles triangle.

5.  ABC is an isosceles triangle with AB = AC. Draw AP ⊥ BC to show that ∠B = ∠C.

Ans. We have AP ⊥ BC

[Given]

          ∴ ∠APB = 90° and APC = 90°

          In ΔABP and ΔACP, we have

          ∠APB = ∠APC

[each = 90°]

          AB = AC

[Given]

          AP = AP

[Common]

          ∴ Using RHS criteria,ΔABP ≌ ΔACP

          ∴ Their corresponding parts are congruent.

          ⇒ ∠B = ∠C

1.  Show that in a right angled triangle, the hypotenuse is the longest side

Ans. Let us consider ΔABC such that ∠B = 90°

          ∴ ∠A + ∠B + ∠C = 180°

          ∴ [∠A + ∠C] + ∠B = 180°

          ⇒ ∠A + ∠C = ∠B

          ∴ ∠B > ∠A and ∠B > ∠C

          ⇒ Side opposite to ∠B is longer than the side opposite to ∠A.

          i.e. AC > BC

...(1)

          Similarly, AC > AB

...(2)

          From (1) and (2), we get AC is the longest side.

          But AC is the hypotenuse of the triangle.

          Thus, hypotenuse is the longest side.

2.  In the adjoining figure, sides AB and AC of ΔABC are extended to points P and Q respectively. Also, ∠PBC < ∠QCB. Show that AC > AB.

Ans. ∠ABC + ∠PBC = 180°

[Linear pair]

            ∠ACB + ∠QCB = 180°

[Linear pair]

            ∴ ∠ABC + ∠PBC = ∠ACB + ∠QCB

            But ∠PBC < ∠QCB

[Given]

            ∴ ∠ABC > ∠ACB

            ⇒ [The side opposite to ∠ABC]

> [The side opposite to ∠ACB]

            ⇒ AC > AB

3.  In the figure, ∠B < ∠A and ∠C < ∠D. Show that AD > BC.

Ans. ∵∠B < ∠A

[Given]

                ⇒ ∠A > ∠B

                ∴ OB > OA

[Side opposite to greater angle is longer]

...(1)

                Similarly,

                OC > OD

...(2)

                From (1) and (2), we have

                [OB + OC] > [OA + OD]

                ⇒ BC > AD

                or AD < BC

4.  AB and CD are respectively the smallest and longest sides of a quadrilateral ABCD (see figure). Show that ∠A > ∠C and ∠B > ∠D.

Ans. Let us join AC.

                Now, in ΔABC,          AB < BC

                [∵AB is the smallest side of quadrilateral ABCD]

                ⇒ BC > AB

∴ [Angle opposite to BC] < [Angle opposite to AB]

                ⇒ ∠BAC > ∠BCA

...(1)

                Again, in ΔACD,

                CD > AD

[∵CD is the longest side of the quadrilateral ABCD]

∴ [Angle opposite to CD] > [Angle opposite to AD]

                ⇒∠CAD > ∠ACD

...(2)

                Adding (1) and (2), we get

                [∠BAC + CAD] > [∠BCA + ∠ACD]

                ⇒ ∠A > ∠C

                Similarly, by joining BD, we have

                ∠B > ∠D

5.  In the figure, PR > PQ and PS bisects ∠QPR. Prove that ∠PSR > ∠PSQ.

Ans. In ΔPAR, PS bisects ∠QPR

[Given]

                ∴ ∠QPS = ∠RPS

                ∴ PR > PQ

[Given]

∴ [Angle opposite to PR] > [Angle opposite to PQ]

                ⇒ ∠PQS > ∠PRS

                ⇒ [∠PQS + ∠QPS] > [∠PRS + ∠RPS]

...(1)

                [∴ ∠QPS = ∠RPS]

                ∴ Exterior ∠PSR = [∠PQS + ∠QPS]

[∴ AN exterior angle is equal to the sum of interior opposite angles]

                And Exterior ∠PSQ = [∠PRS + ∠RPS]

                Now, from (1), we have

                ∠PSR > ∠PSQ

6.  Show that of all line segments drawn from a given point not on it, the perpendicular line segment is the shortest.

Ans. Let us consider the ΔPMN such that ∠M = 90°

                Since, ∠M + ∠N + ∠P = 180°

                [Sum of angles of a triangle]

                ∴ ∠M = 90°

                [∴ PM ⊥ l]

                ⇒ ∠N < ∠M

                ⇒ PM < PN

...(1)

                Similarly, Pm < PN₁

...(2)

                PM < PN₂

...(3)

                From (1), (2) and (3), we have PM is the smallest line segment drawn from P on the line l. Thus, the perpendicular segment is the shortest line segment drawn on a line from a point not on it.

1.  ABC is a triangle. Locate a point in the interior of ΔABC which is equidistant from all the vertices of ΔABC.

Ans. Let us consider a ΔABC.

                Draw ‘l’ the perpendicualar bisector of AB.

                Draw ‘m’ the perpendicular bisector of BC.

                Let the two perpendicular bisectors ‘l’ and ‘m’ meet at O.

                ‘O’ is the required point which is equidistant from A, B and C.

                Note: If we draw a circle with centre ‘O’ and radius OB or OC, then it will pass through A, B and C.

2.  In a triangle, locate a point in its interior which is equidistant from all the sides of the triangle.

Ans. Let us consider a ΔABC.

                Draw ‘l’ the bisector of ∠B.

                Draw ‘m’ the bisector of ∠C.

                Let the two bisectors l and m meet at O. Thus, ‘O’ is the required point which is equidistant from the sides of ΔABC.

                Note: If we draw OM ⊥ BC and draw a circle with O as centre and OM as radius, then the circle will touch the sides of the triangle.

3.  In a huge park, people are concentrated at three points (see figure):

      A: where there are different slides and swings form

      B: near which a man-made lake is situated,

      C: which is near to a large parking and exit.

      Where should an ice cream parlour be set up so that maximum number of persons can approach it?

      Hint: The parlour should be equidistant from A, B and C.

Ans. Let us join A and B, and draw ‘l’ the perpendicular bisector of AB.

            Now, join B and C, and draw ‘m’ the perpendicular bisector of BC. Let the perpendicular bisectors ‘l’ and ‘m’ meet at ‘O’. The point ‘O’ is the required point where the ice cream parlour be set up.

            Note: If we join ‘A’ and ‘C’, and draw the perpendicular bisectors, then it will also meet (or pass through) the point O.

4.  Complete the hexagonal and star shaped Rangolies [see Fig. (i) and (ii)] by filling them with as many equilateral triangles of side 1 cm as you can. Count the number of triangles in each case. Which has more triangles?

Ans. It is an activity. We get the 150 equilateral triangles in the figure (i) and 300 equilateral triangles in the figure (ii).

            ∴ The figure (ii) has more triangles.