Chapter-7 TRIANGLES

1.In quadrilateral ACBD, AC = AD and AB bisect ∠A (see Fig. 7.16). Show that ΔABC≅  ΔABD. What can you say about BC and BD?



Explanation:

It has been given that, AC=AD and AB divides ∠A in two equal parts.

To prove, ΔABC ≅ ΔABD

Proof-

In the given triangle,

(i) AB is common in both.

(ii) ∠CAB = ∠DAB (angle divided in two equal parts)

(iii) AD = AC

So, by Side Angle Side condition (SAS)- ΔABC ≅ ΔABD.

By CPCT rule BC=BD.


2. ABCD is a quadrilateral in which AD = BC and ∠DAB = ∠CBA (see Fig. 7.17). Prove that

(i) ΔABD ≅ ΔBAC

(ii) BD = AC

 (iii) ∠ABD = ∠BAC.



Explanation:

It has been given that, BC=AD and ∠BAD = ∠ABC.

(i) AB is common

⇨ ∠CBA = ∠DAB

⇨ BC = AD

So, by SAS rule ΔABD ≅ ΔBAC (proved)

(ii) As, ΔABD ≅ ΔBAC.

By CPCT condition, AC= BD.

(iii) As, ΔABD ≅ ΔBAC

By CPCT condition, ∠ABD = ∠BAC.


3. AD and BC are equal perpendiculars to a line segment AB (see Fig. 7.18). Show that CD bisects AB.



Explanation:

It has been given that, AD=BC and ∠A = ∠B = 90°.

To prove, OA=OB

In triangles ΔAOD and ΔBOC

(a) AD = BC (given)

(b) ∠A = ∠B (given)

(c) ∠AOD = ∠BOC (opposite angles are equal)

So, ΔAOD ≅ ΔBOC by SAA condition.

By CPCT rule OA=OB (proved)

CD divides AB in two equal parts.


4. l and m are two parallel lines intersected by another pair of parallel lines p and q (see Fig. 7.19). Show that ΔABC ≅ ΔCDA.




Explanation:

It has been given that, l is parallel to m and p is parallel to q.

Proof that, ΔABC ≅ ΔCDA 

Taking ΔABC and ΔCDA we have

(a) ∠DAC = ∠BCA (alternate interior angles)

(b) AC is common in both triangles

(c) ∠DCA = ∠BAC (alternate interior angles)

So, ΔABC ≅ ΔCDA by ASA condition.


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 Fig. 7.20). Show that:

(i) ΔAPB ≅ ΔAQB

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



Explanation:

It has been given that line l bisects ∠A. ∠P = ∠Q=90°

(i) In ΔAPB and ΔAQB 

⇨ AB is common side in both triangle

⇨ ∠BAP = ∠BAQ (given)

⇨ ∠P = ∠Q = 90° (given)

So, by SAA condition ΔAPB ≅ ΔAQB (proved)

(ii) By CPCT rule BQ = BO.


6. In Fig. 7.21, AC = AE, AB = AD and ∠BAD = ∠EAC. Show that BC = DE.



Explanation:

It has been given that CA = EA, AB = AD, and ∠DAB = ∠CAE

To show, BC = DE

⇨ ∠BAD = ∠EAC

Add ∠DAC on LHS and RHS,

⇨ ∠DAC + ∠BAD = ∠DAC + ∠EAC

⇨ ∠BAC = ∠EAD

Taking ΔABC and ΔADE

(a) AB = AD (given)

(b) ∠BAC = ∠EAD

(c) AC = AE (given)

So, by SAS rule ΔABC ≅ ΔADE.

By CPCT rule, 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 Fig. 7.22). Show that

(i) ΔDAP ≅ ΔEBP

(ii) AD = BE



Explanation:

It has been given that, P is mid-point so, AP=BP, ∠DAB = ∠EBA and ∠APE = ∠BPD

(i) In ΔDAP and ΔEBP

⇨ ∠APE = ∠BPD

Add ∠DPE on LHS and RHS

⇨ ∠APE +∠EPD = ∠BPD+∠EPD

⇨ ∠APD = ∠BPE

In ΔAPD and ΔBPE

⇨ ∠DAB = ∠EBA (given)

⇨ AP = BP (given)

⇨ ∠APD = ∠BPE

So, by ASA rule ΔDAP ≅ ΔEBP.

(ii) By CPCT rule BE = AD.


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 Fig. 7.23). Show that:

(i) ΔAMC ≅ ΔBMD

(ii) ∠DBC is a right angle.

(iii) ΔDBC ≅ ΔACB

(iv) CM = AB/2



Explanation:

It has been given that AM=BM, DM=CM, ∠C = 90°

(i) In ΔAMC and ΔBMD:

⇨ AM = BM (given)

⇨ ∠CMA = ∠DMB (opposite angles)

⇨ CM = DM (given)

So, by SAS rule ΔAMC ≅ ΔBMD

(ii) By CPCT rule, ∠MCA = ∠MDB

⇨ So, AC || BD.

⇨ ∠BCA + ∠CBD = 180° (adjacent interior angles in parallel sides)

⇨ 90° +∠CBD = 180°

⇨ ∠CBD = 90°

(iii) In ΔCBD and ΔBCA,

⇨ DB = AC (rule of CPCT)

⇨ ∠ACB = ∠DBC = 90°

⇨ BC is common in both triangles

So, by SAS condition ΔDBC ≅ ΔACB.

(iv) Because, ΔCBD ≅ ΔBCA

⇨ CD = BA

⇨ MD = MB = MC = MA (M is mid-point)

⇨ MD + MC = MB+MA

⇨ MC + MC = AB

⇨ MC = (½) AB (proved)


Exercise 7.2

9.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

Explanation:

It has been given that, AB = AC and ∠OBC and ∠OBE

(i) In triangle ABC

⇨ ∠B = ∠C

⇨ (∠B)/2 = (∠C)/2

⇨ ∠CBO = ∠BCO (given angle is bisected)

⇨ OB = OC (triangle OBC is also isosceles triangle)

(ii) In triangle AOB and triangle AOC,

⇨ AB = AC (given)

⇨ AO = AO (side is common)

⇨ OB = OC (Proved in previous part)

So, by SSS condition ΔAOB ≅ ΔAOC.

By rule of CPCT ∠OAB = OAC

So, OA is bisecting ∠A.


10. In ΔABC, AD is the perpendicular bisector of BC (see Fig. 7.30). Show that ΔABC is an isosceles triangle in which AB = AC.



Explanation:

It has been given that BD=DC and ∠BDA=90°

Proof that, AB = AC

In ΔBDA and ΔCDA,

⇨ BD = DC (given bisector AD to side BC)

⇨ AD (common side in both triangle)

⇨ ∠BDA = ∠CDA = 90°

So, by SAS condition ΔADB ≅ ΔADC.

So, by CPCT condition AB = AC (hence proved)


11. ABC is an isosceles triangle in which altitudes BE and CF are drawn to equal sides AC and AB respectively (see Fig. 7.31). Show that these altitudes are equal.



Explanation:

It has been given that CA = BA and ∠BFC = ∠BEC = 90° 

Proof that, BE = CF

In ΔAEB and ΔAFC

⇨ BA = CA (given)

⇨ ∠A (is common in both triangles)

⇨ ∠BEA = ∠CFA = 90°

So, by SAA condition ΔAEB ≅ ΔAFC

So, By CPCT condition EB = FC (proved)


12. ABC is a triangle in which altitudes BE and CF to sides AC and AB are equal (see Fig. 7.32). Show that

(i) ΔABE ≅ ΔACF

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

Explanation:

It has been given that BE = CF

(i) In ΔEBA and ΔFCA,

⇨ ∠A (angle common in both triangle)

⇨ ∠BEA = ∠CFA = 90°

⇨ EB = FC (given)

So, by AAS condition ΔABE ≅ ΔACF.

(ii) By CPCT rule BA = CA and ΔABC is isosceles.


13. ABC and DBC are two isosceles triangles on the same base BC (see Fig. 7.33). Show that ∠ABD = ∠ACD.



Explanation:

It has been given that, ABC and CBD are isosceles

Proof that, ∠ABD = ∠ACD

⇨ BD = CD (because CBD is isosceles)

⇨ AD (common in both triangles)

⇨ AB = AC (because BAC is isosceles)

So, by SSS condition ΔABD ≅ ΔACD

So, by CPCT condition ∠ABD = ∠ACD (hence proved)


14. ΔABC is an isosceles triangle in which AB = AC. Side BA is produced to D such that AD = AB (see Fig. 7.34). Show that ∠BCD is a right angle.



Explanation:

It has been given, AB = CA = DA

Proof that, ∠BCD =90°

Take triangle ABC,

⇨ AB = CA (given)

⇨ ∠BCA = ∠CBA (because ABC is isosceles)

Take triangle ACD,

⇨ DA = AB (given)

⇨ ∠CDA = ∠DCA (because ACD is isosceles)

Take ΔABC,

⇨ ∠BAC + ∠BCA + ∠ABC = 180°

⇨ ∠BAC + 2∠BCA = 180°

⇨ ∠BAC = 180° – 2∠BCA — (i)

In ΔADC also,

⇨ ∠DAC = 180° – 2∠DCA — (ii)

⇨ ∠BAC + ∠DAC = 180° (DB is line.) —(iii) 

Add (i) and (ii)

∠BAC + ∠DAC = 180° – 2∠BCA+180° – 2∠DCA (use (iii))

⇨ 180° – 360°= – 2∠BCA – 2∠DCA

⇨ (∠BCA+∠DCA)×2 = 180°

⇨ ∠DCB = 180°/2

⇨ ∠DCB = 90° (proved)


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

Explanation:







It has been given that,

⇨ AB = AC and ∠A = 90°

Because, AB = AC

⇨ ∠B = ∠C (isosceles triangle criteria)

⇨ ∠A+∠B+∠C = 180° (interior angles of triangle)

⇨ 90° + ∠C+∠C = 180°

⇨ 2∠C = 90°

⇨ ∠C = 90°/2

⇨ ∠C = 45°

⇨ ∠C = ∠B = 45°


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

Explanation:

In an equilateral triangle, all sides are equal. Let triangle is ABC.

So, AB = BC = CA

⇨ ∠C = ∠A =∠B (angles are equal of the opposite sides)

⇨ ∠C+∠A+∠B = 180° (angles of triangle)

⇨ ∠C+∠C+∠C = 180°

⇨ 3∠C = 180°

⇨ ∠C = 60°

⇨ ∠C = ∠A = ∠B = 60°

Angles in equilateral triangle is 60°.


Exercise 7.3

17. 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 Fig. 7.39). 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.



Explanation:

It has been given that ABC and DBC are isosceles triangles.

(i) In ΔDBA and ΔDCA

⇨ AD is common in both

⇨ BA = CA (CBA isosceles triangle)

⇨ DB = DC (CBD isosceles triangle)

So, by SSS condition ΔABD ≅ ΔACD.


(ii) In triangles PBA and PCA

⇨ PA is common in both triangles

⇨ ∠BAP = ∠CAP (by CPCT rule)

⇨ BA = CA (given)

So, by SAS condition ΔABP ≅ ΔACP


(iii) we know, ΔABD ≅ ΔACD

∠BAP = ∠CAP (by rule of CPCT)

AP divides ∠A (given) - (i)

Now,

⇨ PD is common in both triangle

⇨ DB = DC (CBD isosceles triangle)

⇨ PB = PC (by rule of CPCT)

So, by SSS condition ΔDPB ≅ ΔDPC.

And, ∠PDB = ∠PDC (by rule of CPCT) - (ii)

Analysing (i) and (ii) AP divides ∠D and ∠A.


(iv) ∠DPB = ∠CPD (proved earlier)

⇨ BP = CP - (i)

⇨ ∠DPB +∠CPD = 180° (Straight line BC)

⇨ 2∠DPB = 180°

⇨ ∠DPB = 180°/2

⇨ ∠DPB = 90° - (ii)

Using equation (i) and equation (ii), we get

AP as perpendicular divider of BC.


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

(i) AD bisects BC (ii) AD bisects ∠A.

Explanation:

It has been given that AD is perpendicular to BC and AB=AC as shown below




(i) Take ΔDBA and ΔDCA,

⇨ ∠BDA = ∠CDA = 90° (given)

⇨ AB = AC (also given)

⇨ AD is common in both triangles

So, by rule of RHS ΔDBA ≅ ΔDCA

By CPCT rule, DB = DC or AD divides CB.

(ii) By CPCT rule, ∠DAB = ∠DAC

So, AD divides ∠A.


19. 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 Fig. 7.40). Show that:

(i) ΔABM ≅ ΔPQN

(ii) ΔABC ≅ ΔPQR



Explanation:

It has been given that,

CB = RQ, AB = PQ, MA = NP, BM=QN (AM and PN are median)

(i) In ΔMBA and ΔNQP,

⇨ MA = NP and BA = QP (given)

⇨ BM = QN (median)

So, by SSS rule ΔMBA ≅ ΔNQP

(ii) In ΔABC and ΔPQR

⇨ BA = QP (given)

⇨ CB = RQ (given)

⇨ ∠CBA = ∠RQP (CPCT rule)

So, by SAS rule ΔABC ≅ ΔPQR.


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

Explanation:

It has been given that BE=CF

In ΔCEB and ΔBFC

⇨ ∠BEC = ∠CFB = 90° (because of altitudes)

⇨ CB (common in both triangles)

⇨ BE = CF (given)

So, by RHS rule ΔBEC ≅ ΔCFB.

By CPCT rule ∠B = ∠C 

So, BA = CA (equal angles have equal opposite sides)


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

Explanation:

It has been given that AB = AC

In ΔPBA and ΔPCA

⇨ ∠BPA = ∠CPA = 90° (given altitude)

⇨ AP (common in both triangles)

⇨ AB = AC (given)

So, by RHS rule ΔABP ≅ ΔACP.

So, by CPCT rule ∠C = ∠B.


Exercise 7.4 

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

Explanation:

Triangle ABC is right angle triangle, ∠A=90°.

⇨ ∠A +∠B+∠C = 180° (interior sum of angles in any triangle)

⇨ 90° +∠B+∠C = 180°

⇨ ∠B+∠C = 90°

∠B and ∠C must be less than 90°. So, ∠A is the greatest angle of ABC.

Because ∠A is largest angle it’s opposite side also will be the largest side of this triangle.

The opposite side is BC which is called hypotenuse.


23. In Fig. 7.48, sides AB and AC of ΔABC are extended to points P and Q respectively. Also, ∠PBC < ∠QCB. Show that AC > AB.

Explanation:

It has been given ∠PBC < ∠QCB

⇨ ∠ABC + ∠CBP = 180° (straight line condition)

⇨ ∠ABC = 180° – ∠CBP

⇨ ∠BCA +∠BCQ = 180°

⇨ ∠BCA = 180° – ∠BCQ

⇨ ∠CBP < ∠BCQ (given)

So, ∠ABC > ∠ACB

∠B > ∠C (side opposite to angles will also be greater)

So, AC > AB


24. In Fig. 7.49, ∠B < ∠A and ∠C < ∠D. Show that AD < BC.



Explanation:

⇨ OB > OA (because ∠A > ∠B) —(i)

Similarly, OC > OD —(ii)

Add (i) & (ii)

⇨ OB + OC > OA+OD 

⇨ DA < CB (proved)


25. AB and CD are respectively the smallest and longest sides of a quadrilateral ABCD (see Fig. 7.50).

Show that ∠A > ∠C and ∠B > ∠D.

Explanation:

Take ΔDBA

BA < DA < DB

⇨ ∠BDA < ∠DBA — (i) (angles opposite to the sides)

In ΔDCB

CB < CD < DB

So, we can say that

⇨ ∠CDB < ∠DBC — (ii)

Add (i) & (ii)

⇨ ∠BDA + ∠CDB < ∠DBA + ∠DBC

⇨ ∠CDA < ∠CBA

⇨ ∠D < ∠B

Now take triangle ABC,

⇨ ∠BCA < ∠CAB — (iii) (angles opposite to the sides)

Take ΔADC

⇨ ∠ACD < ∠CAD — (iv)

Add (iii) & (iv)

⇨ ∠BCA + ∠ACD < ∠CAB+∠CAD

⇨ ∠DCB < ∠DAB

⇨ ∠C < ∠A


26. In Fig 7.51, PR > PQ and PS bisect ∠QPR. Prove that ∠PSR > ∠PSQ.

Explanation:

It has been given PQ < PR & PS divides ∠P.

To prove, ∠PSQ < ∠PSR

Now,

⇨ ∠SPQ = ∠SPR — (i) (PS divides ∠P)

⇨ ∠RQP > ∠QRP — (ii) (As PQ < PR so opposite angle to greater side is greater)

⇨ ∠RSP = ∠RQP + ∠SPQ — (iii) (exterior angle equals sum of interior and opposite angles)

⇨ ∠QSP = ∠QRP + ∠SPR — (iv) (exterior angle equals sum of interior and opposite angles)

Add (i) and (ii)

⇨ ∠RQP +∠SPQ > ∠QRP +∠SPR

Using all these equations (i), (ii), (iii) and (iv)

⇨ ∠PSR > ∠PSQ (hence proved)


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

Explanation:

Let us take the line segment “x” and B be the point on it. Draw AB perpendicular to line “x”. Take another point C on line x which is away from B. The figure looks like.




Proof that, AC > AB

Take ΔABC, ∠B = 90° (AB perpendicular to BC)

⇨ ∠A+∠B+∠C = 180° (interior angles property in any triangle)

⇨ ∠A +90°+∠C = 180°

⇨ ∠A+∠C = 90°

As sum of two angles equals 90° then we can surely say that ∠B is the greatest angle 

So, ∠B > ∠A or ∠B > ∠C 

Or AC > AB (side opposites to the angles are also greater)