the sides BA and .CA have been produced such that BA =.CA and CA=AE. prove that DE parallel to BC
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In ΔADE and ΔABCBA = AD [Given]CA=AE [Given]∠1 = ∠ 2 [VOA]ΔADE ≅ ΔABC [SAS]
SO ,
∠ADE = ∠ABC [cpct]∠ADE and ∠ABC are the alternate interior angles for the line segment DE and BC
Hence Proved that DE || BC
SO ,
∠ADE = ∠ABC [cpct]∠ADE and ∠ABC are the alternate interior angles for the line segment DE and BC
Hence Proved that DE || BC
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