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# NCERT Solutions for Class 9th Maths: Chapter 7 Triangles Exercise 7.3

Discussion in 'CBSE Class 9 Maths Help' started by saralas, May 18, 2014.

1. ### saralasMember

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 the given 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.

2. ### saralasMember

(i) In ABD and ACD
AB = AC (given)
BD = CD (given)

(ii) In ABP and ACP
AB = AC (given).
BAP = CAP [from equation (1)]
AP = AP (common)
3. ### saralasMember

(iii) From equation (1)
BAP = CAP
Hence, AP bisect A
Now in BDP and CDP
BD = CD (given)
DP = DP (common)
BP = CP [from equation (2)]

(iv) We have BDP CDP

Now, BPD + CPD = 180o (linear pair angles)

BPD + BPD = 180o

2BPD = 180o [from equation (4)]

BPD = 90o ...(5)

From equations (2) and (5), we can say that AP is perpendicular bisector of BC.