### Exercise 7.1

**Question 1. In quadrilateral ABCD (See figure). AC = AD and AB bisects A. Show that ABC ABD. What can you say about BC and BD?**

**Solution:**

Given: In quadrilateral ABCD, AC = AD and AB bisects A.

To prove: ABC ABD

Proof: In ABC and ABD,

AC = AD [Given]

BAC = BAD [AB bisects A]

AB = AB [Common]

ABC ABD [By SAS congruency]

Thus BC = BD [By C.P.C.T.]

**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

**Solution:**

(i)In ABC and ABD,

BC = AD [Given]

DAB = CBA [Given]

AB = AB [Common]

ABC ABD [By SAS congruency]

Thus AC = BD [By C.P.C.T.]

(ii)Since ABC ABD

AC = BD [By C.P.C.T.]

(iii)Since ABC ABD

ABD = BAC [By C.P.C.T.]

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

**Solution:**

In BOC and AOD,

OBC = OAD = [Given]

BOC = AOD [Vertically Opposite angles]

BC = AD [Given]

BOC AOD [By ASA congruency]

OB = OA and OC = OD [By C.P.C.T.]

**Question 4.and are two parallel lines intersected by another pair of parallel lines and (See figure). Show that ABC CDA.**

**Solution:**

AC being a transversal. [Given]

Therefore DAC = ACB [Alternate angles]

Now [Given]

And AC being a transversal. [Given]

Therefore BAC = ACD [Alternate angles]

Now In ABC and ADC,

ACB = DAC [Proved above]

BAC = ACD [Proved above]

AC = AC [Common]

ABC CDA [By ASA congruency]

**Question 5. Line is the bisector of the angle A and B is any point on BP and BQ are perpendiculars from B to the arms of A. Show that:**

(i) APB AQB

(ii) BP = BQ or P is equidistant from the arms of A (See figure).

**Solution:**

Given: Line bisects A.

BAP = BAQ

(i) In ABP and ABQ,

BAP = BAQ [Given]

BPA = BQA = [Given]

AB = AB [Common]

APB AQB [By ASA congruency]

(ii) Since APB AQB

BP = BQ [By C.P.C.T.]

B is equidistant from the arms of A.

**Question 6. In figure, AC = AB, AB = AD and BAD = EAC. Show that BC = DE.^**

**Solution:**

Given that BAD = EAC

Adding DAC on both sides, we get

BAD + DAC = EAC + DAC

BAC = EAD ……….(i)

Now in ABC and AED,

AB = AD [Given]

AC = AE [Given]

BAC = DAE [From eq. (i)]

ABC ADE [By SAS congruency]

BC = DE [By C.P.C.T.]

**Question 7. AB is a line segment and P is the mid-point. D and E are points on the same side of AB such that BAD = ABE and EPA = DPB. Show that:**

(i) DAF FBP

(ii) AD = BE (See figure)

**Solution:**

Given that EPA = DPB

Adding EPD on both sides, we get

EPA + EPD = DPB + EPD

APD = BPE ……….(i)

Now in APD and BPE,

PAD = PBE [BAD = ABE (given),

PAD = PBE]

AP = PB [P is the mid-point of AB]

APD = BPE [From eq. (i)]

DPA EBP [By ASA congruency]

AD = BE [ By C.P.C.T.]

**Question 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) AMC BMD

(ii) DBC is a right angle.

(iii) DBC ACB

(iv) CM = AB

**Solution:**

(i)In AMC and BMD,

AM = BM [AB is the mid-point of AB]

AMC = BMD [Vertically opposite angles]

CM = DM [Given]

AMC BMD [By SAS congruency]

ACM = BDM ……….(i)

CAM = DBM and AC = BD [By C.P.C.T.]

(ii) For two lines AC and DB and transversal DC, we have,

ACD = BDC [Alternate angles]

AC DB

Now for parallel lines AC and DB and for transversal BC.

DBC = ACB [Alternate angles] ……….(ii)

But ABC is a right angled triangle, right angled at C.

ACB = ……….(iii)

Therefore DBC = [Using eq. (ii) and (iii)]

DBC is a right angle.

(iii) Now in DBC and ABC,

DB = AC [Proved in part (i)]

DBC = ACB = [Proved in part (ii)]

BC = BC [Common]

DBC ACB [By SAS congruency]

(iv) Since DBC ACB [Proved above]

DC = AB

AM + CM = AB

CM + CM = AB [DM = CM]

2CM = AB

CM = AB

### NCERT Solutions for Class 9 Maths Exercise 7.2

**Question 1. In an isosceles triangle ABC, with AB = AC, the bisectors of ∠B and ∠C intersect each other at 0. Join A to 0. Show that**

(i) OB = OC

(ii) AO bisects ∠A**Solution:**

i) in ∆ABC, we have

AB = AC [Given]

∴∠ABC = ∠ACB [Angles opposite to equal sides of a A are equal]

⇒∠ABC = ∠ACB

or ∠OBC = ∠OCB

⇒OC = OB [Sides opposite to equal angles of a ∆ are equal]

(ii) In ∆ABO and ∆ACO, we have

AB = AC [Given]

∠OBA = ∠OCA [ ∵∠B = ∠C]

OB = OC [Proved above]

∆ABO ≅∆ACO [By SAS congruency]

⇒∠OAB = ∠OAC [By C.P.C.T.]

⇒AO bisects ∠A.

**Question 2. In ∆ABC, AD is the perpendicular bisector of BC (see figure). Show that ∆ ABC is an isosceles triangle in which AB = AC.****Solution:**

Since AD is bisector of BC.

∴BD = CD

Now, in ∆ABD and ∆ACD, we have

AD = DA [Common]

∠ADB = ∠ADC [Each 90°]

BD = CD [Proved above]

∴∆ABD ≅∆ACD [By SAS congruency]

⇒AB = AC [By C.P.C.T.]

Thus, ∆ABC is an isosceles triangle.

**Question 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.****Solution:**

∆ABC is an isosceles triangle.

∴AB = AC

⇒∠ACB = ∠ABC [Angles opposite to equal sides of a A are equal]

⇒∠BCE = ∠CBF

Now, in ∆BEC and ∆CFB

∠BCE = ∠CBF [Proved above]

∠BEC = ∠CFB [Each 90°]

BC = CB [Common]

∴∆BEC ≅∆CFB [By AAS congruency]

So, BE = CF [By C.P.C.T.]

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

Show that

(i) ∆ABE ≅∆ACF

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

(i) In ∆ABE and ∆ACE, we have

∠AEB = ∠AFC

[Each 90° as BE ⊥AC and CF ⊥AB]

∠A = ∠A [Common]

BE = CF [Given]

∴∆ABE ≅∆ACF [By AAS congruency]

(ii) Since, ∆ABE ≅∆ACF

∴AB = AC [By C.P.C.T.]

⇒ABC is an isosceles triangle.

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

In ∆ABC, we have

AB = AC [ABC is an isosceles triangle]

∴∠ABC = ∠ACB …(1)

[Angles opposite to equal sides of a ∆ are equal]

Again, in ∆BDC, we have

BD = CD [BDC is an isosceles triangle]

∴∠CBD = ∠BCD …(2)

[Angles opposite to equal sides of a A are equal]

Adding (1) and (2), we have

∠ABC + ∠CBD = ∠ACB + ∠BCD

⇒∠ABD = ∠ACD.

**Question 6. ∆ABC 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.****Solution:**

AB = AC [Given] …(1)

AB = AD [Given] …(2)

From (1) and (2), we have

AC = AD

Now, in ∆ABC, we have

∠ABC + ∠ACB + ∠BAC = 180° [Angle sum property of a A]

⇒2∠ACB + ∠BAC = 180° …(3)

[∠ABC = ∠ACB (Angles opposite to equal sides of a A are equal)]

Similarly, in ∆ACD,

∠ADC + ∠ACD + ∠CAD = 180°

⇒2∠ACD + ∠CAD = 180° …(4)

[∠ADC = ∠ACD (Angles opposite to equal sides of a A are equal)]

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°

⇒∠BCD = = 90°

Thus, ∠BCD = 90°

**Question 7. ABC is a right angled triangle in which ∠A = 90° and AB = AC, find ∠B and ∠C.Solution:In ∆ABC, we have AB = AC [Given]∴Their opposite angles are equal.⇒∠ACB = ∠ABCNow, ∠A + ∠B + ∠C = 180° [Angle sum property of a ∆]⇒90° + ∠B + ∠C = 180° [∠A = 90°(Given)]⇒∠B + ∠C= 180°- 90° = 90°But ∠B = ∠C∠B = ∠C = = 45°Thus, ∠B = 45° and ∠C = 45°**

**Question 8. Show that the angles of an equilateral triangle are 60° each.Solution:In ∆ABC, we haveAB = BC = CA[ABC is an equilateral triangle]AB = BC⇒∠A = ∠C …(1) [Angles opposite to equal sides of a A are equal]Similarly, AC = BC⇒∠A = ∠B …(2)From (1) and (2), we have∠A = ∠B = ∠C = x (say)Since, ∠A + ∠B + ∠C = 180° [Angle sum property of a A]∴x + x + x = 180o⇒3x = 180°⇒x = 60°∴∠A = ∠B = ∠C = 60°Thus, the angles of an equilateral triangle are 60° each.**

**NCERT Solutions for Class 9 Maths Exercise 7.3**

**Question 1. ABC and DBC 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) ABD ACD**

**(ii) ABP ACP**

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

**(iv) AP is the perpendicular bisector of BC.**

**Solution:(i)ABC is an isosceles triangle.**

**AB = AC**

**DBC is an isosceles triangle.**

**BD = CD**

**Now in ABD and ACD,**

**AB = AC [Given]**

**BD = CD [Given]**

**AD = AD [Common]**

**ABD ACD [By SSS congruency]**

**BAD = CAD [By C.P.C.T.] ……….(i)**

**(ii)Now in ABP and ACP,**

**AB = AC [Given]**

**BAD = CAD [From eq. (i)]**

**AP = AP**

**ABP ACP [By SAS congruency]**

**(iii)Since ABP ACP [From part (ii)]**

**BAP = CAP [By C.P.C.T.]**

**AP bisects A.**

**Since ABD ACD [From part (i)]**

**ADB = ADC [By C.P.C.T.] ……….(ii)**

**Now ADB + BDP = [Linear pair] ……….(iii)**

**And ADC + CDP = [Linear pair] ……….(iv)**

**From eq. (iii) and (iv),**

**ADB + BDP = ADC + CDP**

**ADB + BDP = ADB + CDP [Using (ii)]**

**BDP = CDP**

**DP bisects D or AP bisects D.**

**(iv)Since ABP ACP [From part (ii)]**

**BP = PC [By C.P.C.T.] ……….(v)**

**And APB = APC [By C.P.C.T.] …….(vi)**

**Now APB + APC = [Linear pair]**

**APB + APC = [Using eq. (vi)]**

**2APB = **

**APB = **

**AP BC ……….(vii)**

**From eq. (v), we have BP PC and from (vii), we have proved AP B. So, collectively AP is perpendicular bisector of BC.**

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

**(i) AD bisects BC.**

**(ii) AD bisects A.**

**Solution:In ABD and ACD,**

**AB = AC [Given]**

**ADB = ADC = [AD BC]**

**AD = AD [Common]**

**ABD ACD [RHS rule of congruency]**

**BD = DC [By C.P.C.T.]**

**AD bisects BC**

**Also BAD = CAD [By C.P.C.T.]**

**AD bisects A.**

**Question 3. Two sides AB and BC and median AM of one triangle ABC are respectively equal to sides PQ and OR and median PN of ∆PQR (see figure). Show that(i) ∆ABC ≅∆PQR(ii) ∆ABM ≅∆PQNSolution:In ∆ABC, AM is the median.∴BM = BC ……(1)In ∆PQR, PN is the median.∴QN = QR …(2)And BC = QR [Given]⇒BC = QR⇒BM = QN …(3) [From (1) and (2)]**

**(i) In ∆ABM and ∆PQN, we haveAB = PQ , [Given]AM = PN [Given]BM = QN [From (3)]∴∆ABM ≅∆PQN [By SSS congruency]**

**(ii) Since ∆ABM ≅∆PQN⇒∠B = ∠Q …(4) [By C.P.C.T.]Now, in ∆ABC and ∆PQR, we have∠B = ∠Q [From (4)]AB = PQ [Given]BC = QR [Given]∴∆ABC ≅∆PQR [By SAS congruency]**

**Question 4. BE and CF are two equal altitudes of a triangle ABC. Using RHS congruence rule, prove that the triangle ABC is isosceles.Solution:Since BE ⊥AC [Given]∴BEC is a right triangle such that ∠BEC = 90°Similarly, ∠CFB = 90°Now, in right ∆BEC and ∆CFB, we haveBE = CF [Given]BC = CB [Common hypotenuse]∠BEC = ∠CFB [Each 90°]∴∆BEC ≅∆CFB [By RHS congruency]So, ∠BCE = ∠CBF [By C.P.C.T.]or ∠BCA = ∠CBANow, in ∆ABC, ∠BCA = ∠CBA⇒AB = AC [Sides opposite to equal angles of a ∆ are equal]∴ABC is an isosceles triangle.**

**Question 5. ABC is an isosceles triangle with AB = AC. Draw AP ⊥BC to show that ∠B = ∠C.Solution:We have, AP ⊥BC [Given]∠APB = 90° and ∠APC = 90°In ∆ABP and ∆ACP, we have∠APB = ∠APC [Each 90°]AB = AC [Given]AP = AP [Common]∴∆ABP ≅∆ACP [By RHS congruency]So, ∠B = ∠C [By C.P.C.T.]**

**NCERT Solutions for Class 9 Maths Exercise 7.4**

**Question 1. Show that in a right angled triangle, the hypotenuse is the longest side.Solution:Let us consider ∆ABC such that ∠B = 90°∴∠A + ∠B + ∠C = 180°⇒∠A + 90°-+ ∠C = 180°⇒∠A + ∠C = 90°⇒∠A + ∠C = ∠B∴∠B > ∠A and ∠B > ∠C⇒Side opposite to ∠B is longer than the side opposite to ∠Ai.e., AC > BC.Similarly, AC > AB.Therefore, we get AC is the longest side. But AC is the hypotenuse of the triangle. Thus, the hypotenuse is the longest side.**

**Question 2. In figure, sides AB and AC of ∆ABC are extended to points P and Q respectively. Also, ∠PBC < ∠QCB. Show that AC > AB.Solution:∠ABC + ∠PBC = 180° [Linear pair]and ∠ACB + ∠QCB = 180° [Linear pair]But ∠PBC < ∠QCB [Given] ⇒180° – ∠PBC > 180° – ∠QCB⇒∠ABC > ∠ACBThe side opposite to ∠ABC > the side opposite to ∠ACB⇒AC > AB.**

**Question 3. In figure, ∠B <∠A and ∠C <∠D. Show that AD < BC.Solution:Since ∠A > ∠B [Given]∴OB > OA …(1)[Side opposite to greater angle is longer]Similarly, OC > OD …(2)Adding (1) and (2), we haveOB + OC > OA + OD⇒BC > AD**

**Question 4. AB and CD are respectively the smallest and longest sides of a quadrilateral ABCD (see figure). Show that ∠A > ∠C and ∠B >∠D.Solution:Let us join AC.Now, in ∆ABC, AB < BC [∵AB is the smallest side of the quadrilateral ABCD] ⇒BC > AB⇒∠BAC > ∠BCA …(1)[Angle opposite to longer side of A is greater]Again, in ∆ACD, CD > AD[ CD is the longest side of the quadrilateral ABCD]⇒∠CAD > ∠ACD …(2)[Angle opposite to longer side of ∆ is greater]Adding (1) and (2), we get∠BAC + ∠CAD > ∠BCA + ∠ACD⇒∠A > ∠CSimilarly, by joining BD, we have ∠B > ∠D.**

**Question 5. In figure, PR > PQ and PS bisect ∠QPR. Prove that ∠PSR >∠PSQ.Solution:In ∆PQR, PS bisects ∠QPR [Given]∴∠QPS = ∠RPSand PR > PQ [Given]⇒∠PQS > ∠PRS [Angle opposite to longer side of A is greater]⇒∠PQS + ∠QPS > ∠PRS + ∠RPS …(1) [∵∠QPS = ∠RPS]∵Exterior ∠PSR = [∠PQS + ∠QPS]and exterior ∠PSQ = [∠PRS + ∠RPS][An exterior angle is equal to the sum of interior opposite angles]Now, from (1), we have∠PSR = ∠PSQ.**

**Question 6. Show that of all line segments drawn from a given point not on it, the perpendicular line segment is the shortest.Solution:Let us consider the ∆PMN such that ∠M = 90°Since, ∠M + ∠N+ ∠P = 180°[Sum of angles of a triangle is 180°]∵∠M = 90° [PM ⊥l]So, ∠N + ∠P = ∠M⇒∠N < ∠M⇒PM < PN …(1)Similarly, PM < PN1…(2)and 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 line segment is the shortest line segment drawn on a line from a point not on it.**

**NCERT Solutions for Class 9 Maths Exercise 7.5**

**Question 1. ABC is a triangle. Locate a point in the interior of ∆ ABC which is equidistant from all the vertices of ∆ ABC.Solution:Let us consider a ∆ABC.Draw l, the perpendicular 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. The point O is called circumcentre of the triangle.**

**Question 2. In a triangle locate a point in its interior which is equidistant from all the sides of the triangle.Solution:Let us consider a ∆ABC.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 ∆ABC.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. Point O is called incentre of the triangle.**

**Question 3. In a huge park, people are concentrated at three points (see figure)A: where these are different slides and swings for children.B: near which a man-made lake is situated.C: which is near to a large parking and exist.Where should an ice-cream parlor be set? up so that maximum number of persons can approach it?[Hint The parlour should be equidistant from A, B and C.]Solution: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 bisector, then it will also meet (or pass through) the point O.**

**Question 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?Solution:It is an activity.We require 150 equilateral triangles of side 1 cm in the Fig. (i) and 300 equilateral triangles in the Fig. (ii).∴The Fig. (ii) has more triangles.**