Question

In Fig. 8.49, AC = AE, AB = AD and ∆BAD = ∆EAC Prove that BC = DE.​

Answer 1

Step-by-step explanation:

☃️Given -

• AC = AE
• AB = AD
• ∠BAD = ∠EAC

☃️To prove -

• BC = DE

☃️Concept -

• Here we'll use SSS congruency to get the desired answer.

☃️Construction -

• Join DE.

☃️Solution -

In ∆ABC and ∆DEA,

AB = AD (given)

AC = AE (given)

∠BAD = ∠EAC (given)

∠BAD + ∠DAC = ∠EAC + ∠DAC

(common ∠)

∴ ∆ABC ≅ ∆ADE (SSS)

=> BC = DE (CPCT)

Answer 2

∠BAD=∠EAC∠BAD+∠DAC

=∠EAC+∠DAC

∴∠BAC=∠DAE

In △BAC and △DAE

AB=AD (Given)

∠BAC=∠DAE (Proved above)

AC=AE (Given)

∴△BAC≅△DAE (By SAS congruence rule)

BC=DE (By CPCT)