In the figure, BCE and ADE are straight lines. Given that ∠BAD = ∠ACD, prove that

(i)CD bisects ∠ACE,

(ii) BD = AD

Solution:

(i) Let ∠BAD and ∠ACD be x

∠DCB = 180° – x (∠s in the opposite segments)

∠ECB = 180° – (180° – x)

∠ECD = x

Since , ∠ECD = ∠ACD = x

Therefore CD bisects ∠ACE

(ii) ∠DBA = x (∠s in the same segment)

since, ∠DBA = ∠BAD = x (base ∠s of isoceles triangle)

Triangle DBA is an isoceles triangle

therefore BD = DA