1. ΔABC and ΔDBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC (see Fig. 7.39). If AD is extended to intersect BC at P, show that (i) ΔABD ≅ ΔACD (ii) ΔABP ≅ ΔACP (iii) AP bisects ∠A as well as ∠D. (iv) AP is the perpendicular bisector of BC.
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.) ⇒ 2∠BPD = 180° ⇒ ∠BPD = 90° —(ii) Now, from equations (i) and (ii), it can be said that AP is the perpendicular bisector of BC.
