Q. 14.2( 387 Votes )

# 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.

Answer :

It is given in the question that:

are two isosceles triangles

**(i)** In ,

AD = AD (Common)

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 (**SAS**)

**Rule**. 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 (

**SSS**)

**Rule**. 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 = 180^{o} (BC is a straight line)

2∠BPD = 180^{o}

∠BPD = 90^{o} (ii)

From (i) and (ii), we get

AP is the perpendicular bisector of BC

Hence, Proved.Rate this question :

Prove that the angles opposite to equal sides of a triangle are equal

RS Aggarwal & V Aggarwal - MathematicsIf the sides of a triangle are produced in order, then the sum of the three exterior angles so formed is

RD Sharma - MathematicsIn Δ *ABC*, if ∠*A* = 100° *AD* bisects ∠*A* and AD⊥BC. Then, ∠*B* =

D is any point on side AC of a ΔABC with AB = AC. Show that CD < BD.

NCERT Mathematics ExemplarIn a triangle *ABC*, if *AB* = *AC* and AB is produced to *D* such that *BD* = *BC*, find ∠*ACD*: ∠*ADC.*

If the side of a triangle are produced in order, Prove that the sum of the exterior angles so formed is equal to four right angles.

RS Aggarwal & V Aggarwal - Mathematics

In Δ *ABC*, ∠*A*=50° and *BC* is produced to a point *D*. If the bisectors of ∠*ABC* and ∠*ACD* meet at *E*, then ∠*E* =

Compute the value of *x* in each of the following figures:

(i)

(ii)

(iii)

(iv)

RD Sharma - Mathematics

In Δ *PQR*, if *PQ*=*QR* and *L, M* and *N* are the mid-point of the sides *PQ, QR* and *RP* respectively. Prove that *LN*=*MN*.

In a right-angled triangle, one of the acute measures 53°^{.} Find the measure of each angle of the triangle.