Draw the equilateral triangle EBC

Consider Δ

AEB and Δ

AEC, we have:

- AE is common edge (hypothesis) 

- AB = AC (hypothesis) 

- EB = EC (Δ

EBC is a equilateral triangle) 

=> Δ

AEB = Δ

AEC (S-S-S) 

=> EABˆ=EACˆ=20o2=10o

Δ

ABC balance at A => ABCˆ=ACBˆ=180o−20o2=80o

Because Δ

EBC is a equilateral triangle so EBCˆ=ECBˆ=60o

⇒EBAˆ=ECAˆ=80o−60o=20o

Consider Δ

AEC and Δ

CDA, we have:

- AC is common edge (hypothesis) 

- DACˆ=ACEˆ=20o

- AD = EC (= BC) 

=> Δ

AEC = Δ

CDA (S-A-S) 

=> ACDˆ=EACˆ=10o

 

Draw the equilateral triangle EBC

Consider \(\Delta\)AEB and \(\Delta\)AEC, we have:

- AE is common edge (hypothesis) 

- AB = AC (hypothesis) 

- EB = EC (\(\Delta\)EBC is a equilateral triangle) 

=> \(\Delta\)AEB = \(\Delta\)AEC (S-S-S) 

=> \(\widehat{EAB}=\widehat{EAC}=\dfrac{20^o}{2}=10^o\)

\(\Delta\)ABC balance at A => \(\widehat{ABC}=\widehat{ACB}=\dfrac{180^o-20^o}{2}=80^o\)

Because \(\Delta\)EBC is a equilateral triangle so \(\widehat{EBC}=\widehat{ECB}=60^o\)

\(\Rightarrow\widehat{EBA}=\widehat{ECA}=80^o-60^o=20^o\)

Consider \(\Delta\)AEC and \(\Delta\)CDA, we have:

- AC is common edge (hypothesis) 

- \(\widehat{DAC}=\widehat{ACE}=20^o\)

- AD = EC (= BC) 

=> \(\Delta\)AEC = \(\Delta\)CDA (S-A-S) 

=> \(\widehat{ACD}=\widehat{EAC}=10^o\)