exercise 9(a) page: 122 - byju's · concise selina solutions for class 9 maths chapter 9-...
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![Page 1: Exercise 9(A) Page: 122 - Byju's · Concise Selina Solutions for Class 9 Maths Chapter 9- Triangles [Congruency in Triangles] Solution In ΔAOD and ΔBOC, ∠ AOD = ∠ BOC (vertically](https://reader033.vdocuments.site/reader033/viewer/2022043017/5f394e3d8eca8a46355f8b3c/html5/thumbnails/1.jpg)
Concise Selina Solutions for Class 9 Maths Chapter 9-
Triangles [Congruency in Triangles]
Exercise 9(A) Page: 122
1. Which of the following pairs of triangles are congruent? In each case, state the
condition of congruency:
(a) In 𝚫 ABC and 𝚫 DEF, AB=DE, BC=EF and ∠B=∠E.
(b) In 𝚫 ABC and 𝚫 DEF, ∠B=∠E = 90o; AC=DF and BC=EF.
(c) In 𝚫 ABC and 𝚫 QRP, AB=QR, ∠B=∠R and ∠C=∠P.
(d) In 𝚫 ABC and 𝚫 PQR, AB=PQ, AC=PR and BC=QR.
(e) In 𝚫 ABC and 𝚫 PQR, BC=QR, ∠A=90o, ∠C=∠R=40o and ∠Q=50o.
Solution:
![Page 2: Exercise 9(A) Page: 122 - Byju's · Concise Selina Solutions for Class 9 Maths Chapter 9- Triangles [Congruency in Triangles] Solution In ΔAOD and ΔBOC, ∠ AOD = ∠ BOC (vertically](https://reader033.vdocuments.site/reader033/viewer/2022043017/5f394e3d8eca8a46355f8b3c/html5/thumbnails/2.jpg)
Concise Selina Solutions for Class 9 Maths Chapter 9-
Triangles [Congruency in Triangles]
![Page 3: Exercise 9(A) Page: 122 - Byju's · Concise Selina Solutions for Class 9 Maths Chapter 9- Triangles [Congruency in Triangles] Solution In ΔAOD and ΔBOC, ∠ AOD = ∠ BOC (vertically](https://reader033.vdocuments.site/reader033/viewer/2022043017/5f394e3d8eca8a46355f8b3c/html5/thumbnails/3.jpg)
Concise Selina Solutions for Class 9 Maths Chapter 9-
Triangles [Congruency in Triangles]
![Page 4: Exercise 9(A) Page: 122 - Byju's · Concise Selina Solutions for Class 9 Maths Chapter 9- Triangles [Congruency in Triangles] Solution In ΔAOD and ΔBOC, ∠ AOD = ∠ BOC (vertically](https://reader033.vdocuments.site/reader033/viewer/2022043017/5f394e3d8eca8a46355f8b3c/html5/thumbnails/4.jpg)
Concise Selina Solutions for Class 9 Maths Chapter 9-
Triangles [Congruency in Triangles]
2. The given figure shows a circle with centre O. P is mid-point of chord AB.
Show that OP is perpendicular to AB.
Solution
![Page 5: Exercise 9(A) Page: 122 - Byju's · Concise Selina Solutions for Class 9 Maths Chapter 9- Triangles [Congruency in Triangles] Solution In ΔAOD and ΔBOC, ∠ AOD = ∠ BOC (vertically](https://reader033.vdocuments.site/reader033/viewer/2022043017/5f394e3d8eca8a46355f8b3c/html5/thumbnails/5.jpg)
Concise Selina Solutions for Class 9 Maths Chapter 9-
Triangles [Congruency in Triangles]
3. The following figure shows a circle with centre O.
If OP is perpendicular to AB, prove that AP=BP.
Solution
![Page 6: Exercise 9(A) Page: 122 - Byju's · Concise Selina Solutions for Class 9 Maths Chapter 9- Triangles [Congruency in Triangles] Solution In ΔAOD and ΔBOC, ∠ AOD = ∠ BOC (vertically](https://reader033.vdocuments.site/reader033/viewer/2022043017/5f394e3d8eca8a46355f8b3c/html5/thumbnails/6.jpg)
Concise Selina Solutions for Class 9 Maths Chapter 9-
Triangles [Congruency in Triangles]
4. In a triangle ABC, D is mid-point of BC; AD is produced upto E so that DE =
AD. Prove that:
(i) 𝚫 ABD and 𝚫 ECD are congruent.
(ii) AB=EC
(iii) AB is parallel to EC.
Solution
![Page 7: Exercise 9(A) Page: 122 - Byju's · Concise Selina Solutions for Class 9 Maths Chapter 9- Triangles [Congruency in Triangles] Solution In ΔAOD and ΔBOC, ∠ AOD = ∠ BOC (vertically](https://reader033.vdocuments.site/reader033/viewer/2022043017/5f394e3d8eca8a46355f8b3c/html5/thumbnails/7.jpg)
Concise Selina Solutions for Class 9 Maths Chapter 9-
Triangles [Congruency in Triangles]
5. A triangle ABC has ∠B=∠C.
Prove that:
(i) The perpendiculars from the mid-point of BC to AB and AC are equal.
(ii) The perpendiculars form B and C to the opposite sides are equal.
Solution
![Page 8: Exercise 9(A) Page: 122 - Byju's · Concise Selina Solutions for Class 9 Maths Chapter 9- Triangles [Congruency in Triangles] Solution In ΔAOD and ΔBOC, ∠ AOD = ∠ BOC (vertically](https://reader033.vdocuments.site/reader033/viewer/2022043017/5f394e3d8eca8a46355f8b3c/html5/thumbnails/8.jpg)
Concise Selina Solutions for Class 9 Maths Chapter 9-
Triangles [Congruency in Triangles]
![Page 9: Exercise 9(A) Page: 122 - Byju's · Concise Selina Solutions for Class 9 Maths Chapter 9- Triangles [Congruency in Triangles] Solution In ΔAOD and ΔBOC, ∠ AOD = ∠ BOC (vertically](https://reader033.vdocuments.site/reader033/viewer/2022043017/5f394e3d8eca8a46355f8b3c/html5/thumbnails/9.jpg)
Concise Selina Solutions for Class 9 Maths Chapter 9-
Triangles [Congruency in Triangles]
6. The perpendicular bisector of the sides of a triangle AB meet at I.
Prove that: IA=IB=IC.
Solution
![Page 10: Exercise 9(A) Page: 122 - Byju's · Concise Selina Solutions for Class 9 Maths Chapter 9- Triangles [Congruency in Triangles] Solution In ΔAOD and ΔBOC, ∠ AOD = ∠ BOC (vertically](https://reader033.vdocuments.site/reader033/viewer/2022043017/5f394e3d8eca8a46355f8b3c/html5/thumbnails/10.jpg)
Concise Selina Solutions for Class 9 Maths Chapter 9-
Triangles [Congruency in Triangles]
7. A line segment AB is bisected at point P and through point P another line
segment PQ, which is perpendicular to AB, is drawn. Show that: QA=QB.
Solution
![Page 11: Exercise 9(A) Page: 122 - Byju's · Concise Selina Solutions for Class 9 Maths Chapter 9- Triangles [Congruency in Triangles] Solution In ΔAOD and ΔBOC, ∠ AOD = ∠ BOC (vertically](https://reader033.vdocuments.site/reader033/viewer/2022043017/5f394e3d8eca8a46355f8b3c/html5/thumbnails/11.jpg)
Concise Selina Solutions for Class 9 Maths Chapter 9-
Triangles [Congruency in Triangles]
8. If AP bisects angle BAC and M is any point on AP, prove that the
perpendiculars drawn from M to Ab and AC are equal.
Solution
![Page 12: Exercise 9(A) Page: 122 - Byju's · Concise Selina Solutions for Class 9 Maths Chapter 9- Triangles [Congruency in Triangles] Solution In ΔAOD and ΔBOC, ∠ AOD = ∠ BOC (vertically](https://reader033.vdocuments.site/reader033/viewer/2022043017/5f394e3d8eca8a46355f8b3c/html5/thumbnails/12.jpg)
Concise Selina Solutions for Class 9 Maths Chapter 9-
Triangles [Congruency in Triangles]
9. From the given diagram, in which ABCD is a parallelogram, ABL is a line
segment and E is mid-point of BC.
Prove that:
Solution
![Page 13: Exercise 9(A) Page: 122 - Byju's · Concise Selina Solutions for Class 9 Maths Chapter 9- Triangles [Congruency in Triangles] Solution In ΔAOD and ΔBOC, ∠ AOD = ∠ BOC (vertically](https://reader033.vdocuments.site/reader033/viewer/2022043017/5f394e3d8eca8a46355f8b3c/html5/thumbnails/13.jpg)
Concise Selina Solutions for Class 9 Maths Chapter 9-
Triangles [Congruency in Triangles]
10. In the given figure, AB=DB and AC=DC.
If ∠ABD=58o,
∠DBC=(2x-4)o,
∠ACB=y+15o and
∠DCB=63o; find the values of x and y.
Solution
![Page 14: Exercise 9(A) Page: 122 - Byju's · Concise Selina Solutions for Class 9 Maths Chapter 9- Triangles [Congruency in Triangles] Solution In ΔAOD and ΔBOC, ∠ AOD = ∠ BOC (vertically](https://reader033.vdocuments.site/reader033/viewer/2022043017/5f394e3d8eca8a46355f8b3c/html5/thumbnails/14.jpg)
Concise Selina Solutions for Class 9 Maths Chapter 9-
Triangles [Congruency in Triangles]
11. In the given figure: AB/FD,AC//GE and BD=CE; prove that:
(i) BG=DF
(ii) CF=EG
![Page 15: Exercise 9(A) Page: 122 - Byju's · Concise Selina Solutions for Class 9 Maths Chapter 9- Triangles [Congruency in Triangles] Solution In ΔAOD and ΔBOC, ∠ AOD = ∠ BOC (vertically](https://reader033.vdocuments.site/reader033/viewer/2022043017/5f394e3d8eca8a46355f8b3c/html5/thumbnails/15.jpg)
Concise Selina Solutions for Class 9 Maths Chapter 9-
Triangles [Congruency in Triangles]
Solution
12. In a triangle ABC, AB=AC. Show that the altitude AD is median also.
Solution
![Page 16: Exercise 9(A) Page: 122 - Byju's · Concise Selina Solutions for Class 9 Maths Chapter 9- Triangles [Congruency in Triangles] Solution In ΔAOD and ΔBOC, ∠ AOD = ∠ BOC (vertically](https://reader033.vdocuments.site/reader033/viewer/2022043017/5f394e3d8eca8a46355f8b3c/html5/thumbnails/16.jpg)
Concise Selina Solutions for Class 9 Maths Chapter 9-
Triangles [Congruency in Triangles]
13. In the following figure, BL=CM.
Prove that AD is a median of triangle ABC.
Solution
14. In the following figure, AB=AC and AD is perpendicular to BC. BE bisects angle
B and EF is perpendicular to AB.
![Page 17: Exercise 9(A) Page: 122 - Byju's · Concise Selina Solutions for Class 9 Maths Chapter 9- Triangles [Congruency in Triangles] Solution In ΔAOD and ΔBOC, ∠ AOD = ∠ BOC (vertically](https://reader033.vdocuments.site/reader033/viewer/2022043017/5f394e3d8eca8a46355f8b3c/html5/thumbnails/17.jpg)
Concise Selina Solutions for Class 9 Maths Chapter 9-
Triangles [Congruency in Triangles]
Prove that:
(i) BD=CD
(ii) ED=EF
Solution
15. Use the information in the given figure to prove:
(i) AB=FE
(ii) BD=CF
![Page 18: Exercise 9(A) Page: 122 - Byju's · Concise Selina Solutions for Class 9 Maths Chapter 9- Triangles [Congruency in Triangles] Solution In ΔAOD and ΔBOC, ∠ AOD = ∠ BOC (vertically](https://reader033.vdocuments.site/reader033/viewer/2022043017/5f394e3d8eca8a46355f8b3c/html5/thumbnails/18.jpg)
Concise Selina Solutions for Class 9 Maths Chapter 9-
Triangles [Congruency in Triangles]
Solution
![Page 19: Exercise 9(A) Page: 122 - Byju's · Concise Selina Solutions for Class 9 Maths Chapter 9- Triangles [Congruency in Triangles] Solution In ΔAOD and ΔBOC, ∠ AOD = ∠ BOC (vertically](https://reader033.vdocuments.site/reader033/viewer/2022043017/5f394e3d8eca8a46355f8b3c/html5/thumbnails/19.jpg)
Concise Selina Solutions for Class 9 Maths Chapter 9-
Triangles [Congruency in Triangles]
Exercise 9(B) Page: 125
1. On the sides AB and AC of triangle ABC, equilateral triangle ABD and ACE are
drawn. Prove that:
Solution
![Page 20: Exercise 9(A) Page: 122 - Byju's · Concise Selina Solutions for Class 9 Maths Chapter 9- Triangles [Congruency in Triangles] Solution In ΔAOD and ΔBOC, ∠ AOD = ∠ BOC (vertically](https://reader033.vdocuments.site/reader033/viewer/2022043017/5f394e3d8eca8a46355f8b3c/html5/thumbnails/20.jpg)
Concise Selina Solutions for Class 9 Maths Chapter 9-
Triangles [Congruency in Triangles]
2. In the following diagrams, ABCD is a square and APB is an equilateral triangle.
In each case,
Solution
![Page 21: Exercise 9(A) Page: 122 - Byju's · Concise Selina Solutions for Class 9 Maths Chapter 9- Triangles [Congruency in Triangles] Solution In ΔAOD and ΔBOC, ∠ AOD = ∠ BOC (vertically](https://reader033.vdocuments.site/reader033/viewer/2022043017/5f394e3d8eca8a46355f8b3c/html5/thumbnails/21.jpg)
Concise Selina Solutions for Class 9 Maths Chapter 9-
Triangles [Congruency in Triangles]
![Page 22: Exercise 9(A) Page: 122 - Byju's · Concise Selina Solutions for Class 9 Maths Chapter 9- Triangles [Congruency in Triangles] Solution In ΔAOD and ΔBOC, ∠ AOD = ∠ BOC (vertically](https://reader033.vdocuments.site/reader033/viewer/2022043017/5f394e3d8eca8a46355f8b3c/html5/thumbnails/22.jpg)
Concise Selina Solutions for Class 9 Maths Chapter 9-
Triangles [Congruency in Triangles]
![Page 23: Exercise 9(A) Page: 122 - Byju's · Concise Selina Solutions for Class 9 Maths Chapter 9- Triangles [Congruency in Triangles] Solution In ΔAOD and ΔBOC, ∠ AOD = ∠ BOC (vertically](https://reader033.vdocuments.site/reader033/viewer/2022043017/5f394e3d8eca8a46355f8b3c/html5/thumbnails/23.jpg)
Concise Selina Solutions for Class 9 Maths Chapter 9-
Triangles [Congruency in Triangles]
![Page 24: Exercise 9(A) Page: 122 - Byju's · Concise Selina Solutions for Class 9 Maths Chapter 9- Triangles [Congruency in Triangles] Solution In ΔAOD and ΔBOC, ∠ AOD = ∠ BOC (vertically](https://reader033.vdocuments.site/reader033/viewer/2022043017/5f394e3d8eca8a46355f8b3c/html5/thumbnails/24.jpg)
Concise Selina Solutions for Class 9 Maths Chapter 9-
Triangles [Congruency in Triangles]
![Page 25: Exercise 9(A) Page: 122 - Byju's · Concise Selina Solutions for Class 9 Maths Chapter 9- Triangles [Congruency in Triangles] Solution In ΔAOD and ΔBOC, ∠ AOD = ∠ BOC (vertically](https://reader033.vdocuments.site/reader033/viewer/2022043017/5f394e3d8eca8a46355f8b3c/html5/thumbnails/25.jpg)
Concise Selina Solutions for Class 9 Maths Chapter 9-
Triangles [Congruency in Triangles]
3. In the figure, given below, triangle ABC is right-angled triangle at B. ABPQ and
ACRS are squares. Prove that:
(i) 𝚫ACQ and 𝚫ASB are congruent.
(ii) CQ=BS
![Page 26: Exercise 9(A) Page: 122 - Byju's · Concise Selina Solutions for Class 9 Maths Chapter 9- Triangles [Congruency in Triangles] Solution In ΔAOD and ΔBOC, ∠ AOD = ∠ BOC (vertically](https://reader033.vdocuments.site/reader033/viewer/2022043017/5f394e3d8eca8a46355f8b3c/html5/thumbnails/26.jpg)
Concise Selina Solutions for Class 9 Maths Chapter 9-
Triangles [Congruency in Triangles]
Solution
![Page 27: Exercise 9(A) Page: 122 - Byju's · Concise Selina Solutions for Class 9 Maths Chapter 9- Triangles [Congruency in Triangles] Solution In ΔAOD and ΔBOC, ∠ AOD = ∠ BOC (vertically](https://reader033.vdocuments.site/reader033/viewer/2022043017/5f394e3d8eca8a46355f8b3c/html5/thumbnails/27.jpg)
Concise Selina Solutions for Class 9 Maths Chapter 9-
Triangles [Congruency in Triangles]
4. In a 𝚫 ABC, BD is the median to the side AC, BD is produced to E such that
BD=DE.
Prove that: AE is parallel to BC.
Solution
5. In the adjoining figure, QX and RX are the bisectors of the angles Q and R
respectively of the triangle PQR.
If XS⊥QR and XT⊥PQ; prove that:
![Page 28: Exercise 9(A) Page: 122 - Byju's · Concise Selina Solutions for Class 9 Maths Chapter 9- Triangles [Congruency in Triangles] Solution In ΔAOD and ΔBOC, ∠ AOD = ∠ BOC (vertically](https://reader033.vdocuments.site/reader033/viewer/2022043017/5f394e3d8eca8a46355f8b3c/html5/thumbnails/28.jpg)
Concise Selina Solutions for Class 9 Maths Chapter 9-
Triangles [Congruency in Triangles]
Solution
![Page 29: Exercise 9(A) Page: 122 - Byju's · Concise Selina Solutions for Class 9 Maths Chapter 9- Triangles [Congruency in Triangles] Solution In ΔAOD and ΔBOC, ∠ AOD = ∠ BOC (vertically](https://reader033.vdocuments.site/reader033/viewer/2022043017/5f394e3d8eca8a46355f8b3c/html5/thumbnails/29.jpg)
Concise Selina Solutions for Class 9 Maths Chapter 9-
Triangles [Congruency in Triangles]
6. In parallelogram ABCD, the angles A and C are obtuse. Points X and Y are
taken on the diagonal BD such that the angles XAD and YCB are right angles.
Prove that: XA=YC.
Solution
![Page 30: Exercise 9(A) Page: 122 - Byju's · Concise Selina Solutions for Class 9 Maths Chapter 9- Triangles [Congruency in Triangles] Solution In ΔAOD and ΔBOC, ∠ AOD = ∠ BOC (vertically](https://reader033.vdocuments.site/reader033/viewer/2022043017/5f394e3d8eca8a46355f8b3c/html5/thumbnails/30.jpg)
Concise Selina Solutions for Class 9 Maths Chapter 9-
Triangles [Congruency in Triangles]
7. ABCD is a parallelogram. The sides AB and AD are produced to E and F
respectively, such that AB=BE and AD=DF.
Prove that: 𝚫BEC≅ 𝚫DCF
Solution
![Page 31: Exercise 9(A) Page: 122 - Byju's · Concise Selina Solutions for Class 9 Maths Chapter 9- Triangles [Congruency in Triangles] Solution In ΔAOD and ΔBOC, ∠ AOD = ∠ BOC (vertically](https://reader033.vdocuments.site/reader033/viewer/2022043017/5f394e3d8eca8a46355f8b3c/html5/thumbnails/31.jpg)
Concise Selina Solutions for Class 9 Maths Chapter 9-
Triangles [Congruency in Triangles]
8. In the following figures, the sides AB and BC and the median AD of triangle
ABC are respectively equal to the sides PQ and QR and the median PS of the
triangle PQR. Prove that 𝚫ABC and 𝚫PQR are congruent.
Solution
![Page 32: Exercise 9(A) Page: 122 - Byju's · Concise Selina Solutions for Class 9 Maths Chapter 9- Triangles [Congruency in Triangles] Solution In ΔAOD and ΔBOC, ∠ AOD = ∠ BOC (vertically](https://reader033.vdocuments.site/reader033/viewer/2022043017/5f394e3d8eca8a46355f8b3c/html5/thumbnails/32.jpg)
Concise Selina Solutions for Class 9 Maths Chapter 9-
Triangles [Congruency in Triangles]
9. In the following diagram, AP and BQ are equal and parallel to each other.
![Page 33: Exercise 9(A) Page: 122 - Byju's · Concise Selina Solutions for Class 9 Maths Chapter 9- Triangles [Congruency in Triangles] Solution In ΔAOD and ΔBOC, ∠ AOD = ∠ BOC (vertically](https://reader033.vdocuments.site/reader033/viewer/2022043017/5f394e3d8eca8a46355f8b3c/html5/thumbnails/33.jpg)
Concise Selina Solutions for Class 9 Maths Chapter 9-
Triangles [Congruency in Triangles]
Prove that:
Solution
10. In the following figure, OA=OC and AB=BC.
Prove that:
![Page 34: Exercise 9(A) Page: 122 - Byju's · Concise Selina Solutions for Class 9 Maths Chapter 9- Triangles [Congruency in Triangles] Solution In ΔAOD and ΔBOC, ∠ AOD = ∠ BOC (vertically](https://reader033.vdocuments.site/reader033/viewer/2022043017/5f394e3d8eca8a46355f8b3c/html5/thumbnails/34.jpg)
Concise Selina Solutions for Class 9 Maths Chapter 9-
Triangles [Congruency in Triangles]
Solution
11. The following figure shows a triangle ABC in which AB=AC. M is a point on AB
and N is a point on AC such that BM=CN.
![Page 35: Exercise 9(A) Page: 122 - Byju's · Concise Selina Solutions for Class 9 Maths Chapter 9- Triangles [Congruency in Triangles] Solution In ΔAOD and ΔBOC, ∠ AOD = ∠ BOC (vertically](https://reader033.vdocuments.site/reader033/viewer/2022043017/5f394e3d8eca8a46355f8b3c/html5/thumbnails/35.jpg)
Concise Selina Solutions for Class 9 Maths Chapter 9-
Triangles [Congruency in Triangles]
Prove that:
(i) AM=AN
(ii) 𝚫AMC ≅ 𝚫ANB
(iii) BN=CM
(iv) 𝚫BMC ≅ 𝚫CNB
Solution
![Page 36: Exercise 9(A) Page: 122 - Byju's · Concise Selina Solutions for Class 9 Maths Chapter 9- Triangles [Congruency in Triangles] Solution In ΔAOD and ΔBOC, ∠ AOD = ∠ BOC (vertically](https://reader033.vdocuments.site/reader033/viewer/2022043017/5f394e3d8eca8a46355f8b3c/html5/thumbnails/36.jpg)
Concise Selina Solutions for Class 9 Maths Chapter 9-
Triangles [Congruency in Triangles]
12. In a triangle ABC, AB=BC, AD is perpendicular to side BC and CE is
perpendicular to side AB. Prove that:
AD=CE.
Solution
![Page 37: Exercise 9(A) Page: 122 - Byju's · Concise Selina Solutions for Class 9 Maths Chapter 9- Triangles [Congruency in Triangles] Solution In ΔAOD and ΔBOC, ∠ AOD = ∠ BOC (vertically](https://reader033.vdocuments.site/reader033/viewer/2022043017/5f394e3d8eca8a46355f8b3c/html5/thumbnails/37.jpg)
Concise Selina Solutions for Class 9 Maths Chapter 9-
Triangles [Congruency in Triangles]
13. PQRS is a parallelogram. L and M are points on PQ and SR respectively such
that PL=MR. Show that LM and QS bisect each other.
Solution
14. In the following figure, ABC is an equilateral triangle in which P is parallel to
AC. Side AC is produced up to point R so that CR=BP. Prove that QR bisects
PC.
Solution
![Page 38: Exercise 9(A) Page: 122 - Byju's · Concise Selina Solutions for Class 9 Maths Chapter 9- Triangles [Congruency in Triangles] Solution In ΔAOD and ΔBOC, ∠ AOD = ∠ BOC (vertically](https://reader033.vdocuments.site/reader033/viewer/2022043017/5f394e3d8eca8a46355f8b3c/html5/thumbnails/38.jpg)
Concise Selina Solutions for Class 9 Maths Chapter 9-
Triangles [Congruency in Triangles]
15. In the following figure ∠A=∠C and AB=BC.
![Page 39: Exercise 9(A) Page: 122 - Byju's · Concise Selina Solutions for Class 9 Maths Chapter 9- Triangles [Congruency in Triangles] Solution In ΔAOD and ΔBOC, ∠ AOD = ∠ BOC (vertically](https://reader033.vdocuments.site/reader033/viewer/2022043017/5f394e3d8eca8a46355f8b3c/html5/thumbnails/39.jpg)
Concise Selina Solutions for Class 9 Maths Chapter 9-
Triangles [Congruency in Triangles]
Prove that: 𝚫ABD≅ 𝚫CBE
Solution
In triangles AOE and COD,
∠A = ∠C (given)
∠AOE = ∠COD (vertically opposite angles)
∴ ∠A + ∠AOE = ∠C + ∠COD
⇒180° - ∠AEO = 180° - ∠CDO
⇒ ∠AEO = ∠ CDO ….(i)
Now, ∠AEO + ∠OEB = 180° (linear pair)
And, ∠CDO + ∠ODB = 180° (linear pair)
∴ ∠AEO + ∠OEB = ∠CDO + ∠ODB
⇒ ∠OEB = ∠ODB [Using (i)]
⇒ ∠CEB = ∠ADB ….(ii)
Now, in ΔABD and ΔCBE,
∠A = ∠C (given)
∠ADB = ∠CEB [From (ii)]
AB = BC (given)
⇒ ΔABD ≅ ΔCBE (by AAS congruence criterion)
16. AD and BC are equal perpendiculars to a line segment AB. If AD and BC are
on different sides of AB prove that CD bisects AB.
![Page 40: Exercise 9(A) Page: 122 - Byju's · Concise Selina Solutions for Class 9 Maths Chapter 9- Triangles [Congruency in Triangles] Solution In ΔAOD and ΔBOC, ∠ AOD = ∠ BOC (vertically](https://reader033.vdocuments.site/reader033/viewer/2022043017/5f394e3d8eca8a46355f8b3c/html5/thumbnails/40.jpg)
Concise Selina Solutions for Class 9 Maths Chapter 9-
Triangles [Congruency in Triangles]
Solution
In ΔAOD and ΔBOC,
∠ AOD = ∠ BOC (vertically opposite angles)
∠ DAO = ∠ CBO (each 90°)
AD = BC (given)
∴ ΔAOD ≅ ΔBOC (by AAS congruence criterion)
⇒ AO = BO [cpct]
⇒ O is the mid-point of AB.
Hence, CD bisects AB.
17. In 𝚫ABC, AB=AC and the bisectors of angles B and C intersect at point O.
Prove that:
(i) BO=CO
(ii) AO bisects angle BAC.
Solution
In ΔABC,
AB = AC
⇒ ∠B = ∠C (angles opposite to equal sides are equal)
![Page 41: Exercise 9(A) Page: 122 - Byju's · Concise Selina Solutions for Class 9 Maths Chapter 9- Triangles [Congruency in Triangles] Solution In ΔAOD and ΔBOC, ∠ AOD = ∠ BOC (vertically](https://reader033.vdocuments.site/reader033/viewer/2022043017/5f394e3d8eca8a46355f8b3c/html5/thumbnails/41.jpg)
Concise Selina Solutions for Class 9 Maths Chapter 9-
Triangles [Congruency in Triangles]
18. In the following figure, AB=EF, BC=DE and ∠B=∠E=90o.
Prove that AD=FC.
Solution
![Page 42: Exercise 9(A) Page: 122 - Byju's · Concise Selina Solutions for Class 9 Maths Chapter 9- Triangles [Congruency in Triangles] Solution In ΔAOD and ΔBOC, ∠ AOD = ∠ BOC (vertically](https://reader033.vdocuments.site/reader033/viewer/2022043017/5f394e3d8eca8a46355f8b3c/html5/thumbnails/42.jpg)
Concise Selina Solutions for Class 9 Maths Chapter 9-
Triangles [Congruency in Triangles]
19. A point O is taken inside a rhombus ABCD such that its distance from the
vertices B and D are equal. Show that AOC is a straight line.
Solution
20. In quadrilateral ABCD, AD=BC and BD=CA. Prove that:
![Page 43: Exercise 9(A) Page: 122 - Byju's · Concise Selina Solutions for Class 9 Maths Chapter 9- Triangles [Congruency in Triangles] Solution In ΔAOD and ΔBOC, ∠ AOD = ∠ BOC (vertically](https://reader033.vdocuments.site/reader033/viewer/2022043017/5f394e3d8eca8a46355f8b3c/html5/thumbnails/43.jpg)
Concise Selina Solutions for Class 9 Maths Chapter 9-
Triangles [Congruency in Triangles]
(i) ∠ADB = ∠BCA
(ii) ∠DAB = ∠CBA
Solution
In ΔABD and ΔBAC,
AD = BC (given)
BD = CA (given)
AB = AB (common)
∴ ΔABD ≅ ΔBAC (by SSS congruence criterion)
⇒ ∠ADB = ∠BCA and ∠DAB = ∠CBA (cpct)