In the adjoining figure, ABC and 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) △ABD ≅ △ACD

(ii) △ABP ≅ △ACP

(iii) AP bisects ∠A as well as ∠D

(iv) AP is the perpendicular bisector of BC.

(i) AB = AC (as ABC is an isosceles triangle on base BC)

DB = DC (as DBC is an isosceles triangle on base BC)

AD = AD (Common)

∴ △ABD ≅ △ACD by SSS axiom.

Hence, proved that △ABD ≅ △ACD.

(ii) AB = AC (as ABC is an isosceles triangle on base BC)

AP = AP (Common)

∠ABP = ∠ACP (As angles opposite to equal sides are equal.)

∴ △ABP ≅ △ACP by AAS axiom.

Hence, proved that △ABP ≅ △ACP.

(iii) We know that,

△ABP ≅ △ACP.

As corresponding parts of congruent triangle are equal.

∴ ∠BAP = ∠CAP ……(i)

also ∠APB = ∠APC.

△DPB ≅ △DPC (By SAS axiom)

As corresponding parts of congruent triangle are equal.

∴ ∠BDP = ∠CDP ……(ii)

From (i) and (ii) we can say that AP bisects ∠A as well as ∠D.

Hence, proved that AP bisects ∠A as well as ∠D.

(iv) Since, ∠BPA = ∠CPA.

From figure,

∠BPA + ∠CPA = 180°

2∠BPA = 180°

∠BPA = 90°.

Hence, proved that AP is the perpendicular bisector of BC.