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Answer with explanation</u>
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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.
