Question in pic. Plz open the pic for the question ..corect gets branliest

Given: In Δ ABC and ΔAEC,

AB=BC and AD=CD

i) In ΔADB and CBD, we have

AD = DC        [given]

AB=BC           [given]

DB= DB           [given]

⇒ ΔADB ≅ΔCDB   [By SSS congruence rule]

⇒ ∠ADB ≅∠CDB       ...(i)           [Corresponding parts of congruent triangles are congruent]

Since AC is a straight line,

∠ADB+∠CDB = 180°   [Linear pair]

⇒∠ADB+∠ADB=180°   [from (i)]

⇒2 ∠ADB=180°  

⇒∠ADB=90°  =∠CDB

Also ∠ADB+∠ADE=180°   [Linear pair]

⇒∠ADE=180°-∠ADB  = 180°-90°

⇒∠ADE=90°, i.e. ∠ADE is a right triangle.

Similarly, ∠CDB+∠CDE=180°

⇒∠CDE=90°

ii) Now, in ΔADE and CDE

AD= CD   [given]

ED=ED   [Common]

∠ADE= ∠CDE = 90°

⇒ΔADE ≅ CDE   [By SAS congruence rule ]

⇒AE=EC   [Corresponding parts of congruent triangles are congruent]

Hence proved.

0.84 dang lotta people are looking for this lol or if you round up its 1 but the overall answer to that is 0.84

step-by-step explanation: