Given :
Δ ABC and Δ DBC are isosceles triangles on the same base BC.
∴ AB = AC and DB = DC
(i) In Δ ABD and Δ ACD,
⇒ AB = AC (Equal sides of isosceles Δ ABC)
⇒ BD = CD (Equal sides of isosceles Δ DBC)
⇒ AD = AD (Common)
∴ Δ ABD ≅ Δ ACD (By S.S.S. congruence rule)
Hence, proved that Δ ABD ≅ Δ ACD.
(ii) Since,
Δ ABD ≅ Δ ACD.
We know that,
Corresponding parts of congruent triangles are equal.
⇒ ∠BAD = ∠CAD ........(1)
From figure,
⇒ ∠BAD = ∠BAP and ∠CAD = ∠CAP
Substituting above value in equation (1), we get :
⇒ ∠BAP = ∠CAP ..........(2)
In Δ ABP and Δ ACP,
⇒ AB = AC (Equal sides of isosceles Δ ABC)
⇒ ∠BAP = ∠CAP [From equation (2)]
⇒ AP = AP (Common)
∴ Δ ABP ≅ Δ ACP (By S.A.S. congruence rule)
Hence, proved that Δ ABP ≅ Δ ACP.
(iii) Since,
Δ ABD ≅ Δ ACD
∴ ∠ADB = ∠ADC (By C.P.C.T.) ..........(3)
∴ ∠BAP = ∠CAP [From equation (2)]
∴ AP is the angle bisector of ∠A.
From equation (3),
⇒ ∠ADB = ∠ADC
⇒ 180° - ∠ADB = 180° - ∠ADC
⇒ ∠BDP = ∠CDP ......(4)
∴ AP is the bisector of ∠D
Hence, proved that AP bisects ∠A as well as ∠D.
(iv) In Δ BDP and Δ CDP,
⇒ DP = DP (Common side)
⇒ ∠BDP = ∠CDP [From equation (4)]
⇒ DB = DC (Equal sides of isosceles Δ DBC)
∴ Δ BDP ≅ Δ CDP (By S.A.S. congruence rule)
∴ ∠BPD = ∠CPD (By C.P.C.T.) .......(5)
From figure,
⇒ ∠BPD + ∠CPD = 180° (Linear pair)
⇒ ∠BPD + ∠BPD = 180° [From Equation (5)]
⇒ 2∠BPD = 180°
⇒ ∠BPD = 90° .......(6)
We know that,
⇒ BP = CP [Proved above]
Hence, proved that AP is the perpendicular bisector of BC.