Page 5 - ln1

∴ ΔABP ≅ ΔACP (By SAS congruence rule)
∴ BP = CP (By CPCT) … (2)

(iii) From Equation (1),
∠BAP = ∠CAP
Hence, AP bisects ∠A.
In ΔBDP and ΔCDP,
BD = CD (Given)
DP = DP (Common)
BP = CP [From equation (2)]
∴ ΔBDP ≅ ΔCDP (By SSS Congruence rule)
∴∠BDP = ∠CDP (By CPCT) … (3)
Hence, AP bisects ∠D.

(iv) ΔBDP ≅ ΔCDP
∴∠BPD = ∠CPD (By CPCT) …. (4)
o
∠BPD + ∠CPD = 180 (Linear pair angles)
o
∠BPD + ∠BPD = 180
o
2∠BPD = 180 [From Equation (4)]
o
∠BPD = 90 … (5)
From Equations (2) and (5), it can be said that AP is the perpendicular bisector of BC.

Question 2
Two sides AB and BC and median AM of one triangle ABC are respectively equal
to sides PQ and QR and median PN of ΔPQR (see the given figure). Show that:
(i) ΔABM ≅ ΔPQN
(ii) ΔABC ≅ ΔPQR

(i) In ΔABC, AM is the median to BC.
∴ BM = ½ BC
In ΔPQR, PN is the median to QR.

∴ QN =1/2 QR

However, BC = QR
∴½ BC = ½ QR
∴ BM = QN … (1)
In ΔABM and ΔPQN,
AB = PQ (Given)
BM = QN [From Equation (1)]
AM = PN (Given)
ΔABM ≅ ΔPQN (By SSS congruence rule)
∠ABM = ∠PQN (By CPCT)
∠ABC = ∠PQR … (2)
(ii) In ΔABC and ΔPQR,