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Install NowNCERT Solutions for Class 9 Maths Exercise 7.3 book solutions are available in PDF format for free download. These ncert book chapter wise questions and answers are very helpful for CBSE board exam. CBSE recommends NCERT books and most of the questions in CBSE exam are asked from NCERT text books. Class 9 Maths chapter wise NCERT solution for Maths Book for all the chapters can be downloaded from our website and myCBSEguide mobile app for free.
NCERT solutions for Class 9 Maths Triangles Download as PDF
NCERT Solutions for Class 9 Mathematics Triangles
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.
Ans. (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)]
2
APB =
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.
NCERT Solutions for Class 9 Maths Exercise 7.3
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. 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.
NCERT Solutions for Class 9 Maths Exercise 7.3
3. Two sides AB and BC and median AM of the triangle ABC are respectively equal to side PQ and QR and median PN of
PQR (See figure). Show that:
(i) ABM
PQN
(ii) ABC
PQR
Ans. AM is the median of ABC.
BM = MC =
BC ……….(i)
PN is the median of PQR.
QN = NR =
QR ……….(ii)
Now BC = QR [Given] BC =
QR
BM = QN ……….(iii)
(i) Now in ABM and
PQN,
AB = PQ [Given]
AM = PN [Given]
BM = QN [From eq. (iii)]
ABM
PQN [By SSS congruency]
B =
Q [By C.P.C.T.] ……….(iv)
(ii) In ABC and
PQR,
AB = PQ [Given]
B =
Q [Prove above]
BC = QR [Given]
ABC
PQR [By SAS congruency]
NCERT Solutions for Class 9 Maths Exercise 7.3
4. BE and CF are two equal altitudes of a triangle ABC. Using RHS congruence rule, prove that the triangle ABC is isosceles.
Ans. In BEC and
CFB,
BEC =
CFB [Each
]
BC = BC [Common]
BE = CF [Given]
BEC
CFB [RHS congruency]
EC = FB [By C.P.C.T.] …..(i)
Now In AEB and
AFC
AEB =
AFC [Each
]
A =
A [Common]
BE = CF [Given]
AEB
AFC [ASA congruency]
AE = AF [By C.P.C.T.] …………(ii)
Adding eq. (i) and (ii), we get,
EC + AE = FB + AF
AB = AC
ABC is an isosceles triangle.
NCERT Solutions for Class 9 Maths Exercise 7.3
5. ABC is an isosceles triangles with AB = AC. Draw AP
BC and show that
B =
C.
Ans. Given: ABC is an isosceles triangle in which AB = AC
To prove: B =
C
Construction: Draw AP BC
Proof: In ABP and
ACP
APB =
APC =
[By construction]
AB = AC [Given]
AP = AP [Common]
ABP
ACP [RHS congruency]
B =
C [By C.P.C.T.]
NCERT Solutions for Class 9 Maths Exercise 7.3
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