Question

In the given figure AD=AE D and E are points on BC such that BD=EC.Prove that AB=AC

Answer 1

Answer:

Given,

AD=AE  where, D and E are points on BC,

Such that BD = EC,

To prove :

AB = AC,

Proof :

∵ AD=AE

⇒ ∠ADE = ∠AED

⇒ 180° -  ∠ADE =  180° -  ∠AED

⇒ ∠ADB = ∠AEC,

BD = EC ( given )

By SAS postulate of congruence,

Δ ABD ≅ Δ ACE

By CPCT,

AB = AC

Hence, proved.....

Answer 2

Given:

In the given figure AD=AE D and E are points on BC such that BD=EC.

To prove:

AB=AC.

Proof:

In triangle ADE,

$AD=AE$             [Given]

$\angle ADE=\angle AED$       [Base angles of an isosceles triangle are equal]

Now,

$180^\circ-\angle ADE=180^\circ-\angle AED$         [Linear pair]

$\angle ADB=\angle AEC$         ...(i)

In triangle ABD and ACE,

$AD=AE$              [Given]

$\angle ADB=\angle AEC$             [Using (i)]

$BD=EC$              [Given]

$BD=CE$

Since, two corresponding sides and there included angles are equal, therefore, the triangles are congruent be SAS postulate.

$\Delta ABD\cong \Delta ACE$

We know that corresponding parts of congruent triangles are congruent (CPCTC).

$AB=AC$            [CPCTC]

Hence proved.