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 Figure). 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)
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)
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.