Question

Please help me in solving this question

Answer 1

Given: ∆ABC and ∆DBC are two isosceles triangles on the same base BC.

To prove:





(i) Δ ABD ≅ Δ ACD

(ii) Δ ABE ≅ Δ ACE

(iii) AE bisects ∠ A as well as ∠ D.

(iv) AE is the perpendicular bisector of BC







(i)


Proof:


In ∆ ABC,

AC = AB  (∆ABC is an isosceles triangle)

⇒∠ABC = ∠ACB    … (1)  (Equal sides have equal angles opposite to them)

In ∆ DBC,

DC = DB  (∆DBC is an isosceles triangle)

⇒ ∠DBC = ∠DCB  … (2)  (Equal sides have equal angles opposite to them)

Adding (1) and (2), we get

∠ABC = ∠DBC = ∠ACB + ∠DCB

⇒ ∠ABD = ∠ACD




(ii)In ΔABE and Δ ACE

           AB = AC [given]

       ∠ BAE = ∠CAE [proved above]

             AE = AE [common]

   By SAS Congruence Criterion Rule

             Δ ABE ≅ Δ ACE

                   BE = CE [CPCT] … 2

                    ∠AEB = ∠AEC [CPCT]

(iii)      ∠ BAE = ∠CAE [From eq. 1]

                Hence, AE bisects ∠ A.

 Now, In Δ BDE and Δ CDE

BD = CD [given]

      BE = CE [given]

      DE = DE [common]

By SSS Congruence Criterion Rule

      Δ BDE ≅ Δ CDE

    ∠ BDE = ∠CDE [CPCT]

AE bisects ∠ D.

(iv) AE stands on B

∠AEB + ∠AEC = 1800

∠AEB +∠AEB = 1800[proved above]

    ∠AEB = 1800  /2

 ∠AEB = 900

AE is the perpendicular bisector of BC