Question

In figure it is given that AB=BC and AD=EC prove that​

Answer 1

$\huge \pink {answer}$

In ΔABC,

AB = BC (given)

⇒ ∠BCA = ∠BAC .(Angles op-posite to eq-ual sides are eq-ual)

⇒ ∠BCD = ∠BAE ….(i)

Given, AD = EC

⇒ AD + DE = EC + DE .(Ad-ding DE on both sides)

⇒ AE = CD ..(ii)

Now, in triangl-es ABE and CBD,

AB = BC ..(given)

∠BAE = ∠BCD ....[From (i)]

AE = CD ..[From (ii)]

⇒ ΔABE ≅ ΔCBD

⇒ BE = BD ..(cpct)

Answer 2

$\huge \red {AnSWER}$

In ΔABC,

AB = BC (given)

⇒ ∠BCA = ∠BAC .(Angles op-posite to eq-ual sides are eq-ual)

⇒ ∠BCD = ∠BAE ….(i)

Given, AD = EC

⇒ AD + DE = EC + DE .(Ad-ding DE on both sides)

⇒ AE = CD ..(ii)

Now, in triangl-es ABE and CBD,

AB = BC ..(given)

∠BAE = ∠BCD ....[From (i)]

AE = CD ..[From (ii)]

⇒ ΔABE ≅ ΔCBD

⇒ BE = BD ..(cpct)