# 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 (see Fig. 7.39). 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. It is given in the question that: are two isosceles triangles (i) In ,

AB = AC (Triangle ABC is isosceles)

BD = CD (Triangle DBC is isosceles)

Three sides equal (SSS)Definition: Triangles are congruent if all three sides in one triangle are congruent to the corresponding sides in the other.

Therefore,

By SSS axiom, (ii) In AP = AP (Common)

PAB = PAC (By c.p.c.t)

AB = AC (Triangle ABC is isosceles)

Side-Angle-Side (SASRule. Side-Angle-Side is a rule used to prove whether a given set of triangles are congruent. If two sides and the included angle of one triangle are equal to two sides and included angle of another triangle, then the triangles are congruent

Therefore,

By SAS axiom, (iii) PAB = PAC (By c.p.c.t)

AP bisects A (i)

Also,

In PD = PD (Common)

BD = CD (Triangle DBC is isosceles)

BP = CP ( so by c.p.c.t)

Side-Side-Side (SSSRule. Side-Side-Side is a rule used to prove whether a given set of triangles are congruent. The SSS rule states that: If three sides of one triangle are equal to three sides of another triangle, then the triangles are congruent.

Therefore,

By SSS axiom,  BDP = CDP (By c.p.c.t) (ii)

By (i) and (ii), we can say that AP bisects A as well as D

(iv) BPD = CPD (By c.p.c.t)

And,

BP = CP (i)

Also,

BPD + CPD = 180o (BC is a straight line)

2BPD = 180o

BPD = 90o (ii)

From (i) and (ii), we get

AP is the perpendicular bisector of BC

Hence, Proved.

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