Chapter 7 ma ex 7.3 ma …

In the above question, it is given that ΔABC and ΔDBC are two isosceles triangles. (i) ΔABD and ΔACD are similar by SSS congruency because: AD = AD (It is the common arm) AB = AC (Since ΔABC is isosceles) BD = CD (Since ΔDBC is isosceles) ∴ ΔABD ΔACD. (ii) ΔABP and ΔACP are similar as: AP = AP (It is the common side) PAB = PAC (by CPCT since ΔABD ΔACD) AB = AC (Since ΔABC is isosceles) So, ΔABP ΔACP by SAS congruency condition. (iii) PAB = PAC by CPCT as ΔABD ΔACD. AP bisects A. — (i) Also, ΔBPD and ΔCPD are similar by SSS congruency as PD = PD (It is the common side) BD = CD (Since ΔDBC is isosceles.) BP = CP (by CPCT as ΔABP ΔACP) So, ΔBPD ΔCPD. Thus, BDP = CDP by CPCT. — (ii) Now by comparing (i) and (ii) it can be said that AP bisects A as well as D. (iv) BPD = CPD (by CPCT as ΔBPD ΔCPD) and BP = CP — (i) also,BPD +CPD = 180° (Since BC is a straight line.) ⇒ 2BPD = 180° ⇒ BPD = 90° —(ii) Now, from equations (i) and (ii), it can be said that AP is the perpendicular bisector of BC.