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Question

ΔABC and ΔDBC are two isosceles I 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) ΔABDΔACD
(ii) ΔABPΔACP
(iii) AP bisects A as well as D
(iv) AP is the perpendicular bisector of BC.
1878490_a0dd796b88c141828de7e83fe46e809f.png

Solution
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(i) In Δs ABD and ACD, we have
AB=AC (given)
BD=DC (given)
and AD=AD (common)
ΔABDΔACD
(by SSS congruency rule)
(ii) In Δs ABP and ACP, we have
AB=AC (given)
BAP=CAP
and AP=AP (common)
[ΔABDΔACD,BAD=CAD,BAP=CAP]
ΔABPΔACP
(by SAS congruency rule)
(iii) We have already proved in (i) that
ΔABDΔCAD
BAP=CAP
AP bisects A i.e. AP is the bisector of A.
In Δs BDP and CDP, we have
BD=CD (given)
BP=CP[ΔABP=ΔACP]
and DP=DP (common)
ΔBDPΔCDP
(by SSS congruency rule)
BDP=CDP
DP is the bisector of D.
Hence, AP is the bisector of A as well as D.
(iv) In (iii), we have proved that
ΔBDPΔCDP
BP=CP and BPD=CPD=900.
BPD and CPD form a linear pair
DP is the perpendicular bisector of BC
Hence, AP is the perpendicular bisector of BC

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1878490_a0dd796b88c141828de7e83fe46e809f.png
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