. 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 figure). If AD is extended to intersect BC
at P, show that
(i) AABD = AACD
(ii) AABP = AACP
(iii) AP bisects ZA as well as D.
(iv) AP is the perpendicular bisector of BC.
Answers
Correct Question :-
ΔABC and ΔDBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC. If AD is extended to intersect BC at P, show that
(i) ΔABD ≅ ΔACD
(ii) ΔABP ≅ ΔACP
(iii) AP bisects ZA as well as D.
(iv) AP is the perpendicular bisector of BC.
Given :-
ΔABC and ΔDBC are two isosceles triangles in which AB=AC and BD=DC.
To Find :-
ΔABD ≅ ΔACD
ΔABP ≅ ΔACP
AP bisects ZA as well as D.
AP is the perpendicular bisector of BC.
Solution :-
Given that,
ΔABC and ΔDBC are two isosceles triangles.
(i) ΔABD ≅ ΔACD
ΔABD and ΔACD are similar by SSS congruency.
AD = AD
Since ΔABC is isosceles,
AB = AC
Since ΔDBC is isosceles,
BD = CD
Therefore, ΔABD ≅ ΔACD
(ii) ΔABP ≅ ΔACP
ΔABP and ΔACP are similar.
Common side,
AP = AP
By CPCT since ΔABD ΔACD,
PAB = PAC
Since ΔABC is isosceles,
Therefore, ΔABP ΔACP by SAS congruency.
(iii) AP bisects ZA as well as D
PAB = PAC by CPCT as ΔABD ΔACD
AP bisects A
ΔBPD and ΔCPD are similar by SSS congruency
Common side,
PD = PD
Since ΔDBC is isosceles,
BD = CD
By CPCT as ΔABP ΔACP,
BP = CP
Thus, ΔBPD ΔCPD
BDP = CDP by CPCT.
AP bisects A as well as D.
(iv) AP is the perpendicular bisector of BC.
By CPCT as ΔBPD ΔCPD,
BPD = CPD
BP = CP
Since BC is a straight line,
BPD + CPD = 180°
2BPD = 180°
BPD = 90°
Therefore, AP is the perpendicular bisector of BC.