Δ ABC and Δ DBC 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) Δ ABD ≅ Δ ACD

(ii) Δ ABP ≅ Δ ACP

(iii) AP bisects ∠A as well as ∠D

(iv) AP is the perpendicular bisector of BC

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.