Chapter 7 ma ex 7.3 ma q1

1.	Chapter 7 ma ex 7.3 ma q1
Answer» 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 asPD = 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 thatAP is the perpendicular bisector of BC.

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