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(i) /B +/C+/D =180°(sum of all angles of triangle)
/B +25°+100°=180°
/B +125°=180°
/B =180°-125°
/B =55°
(ii) if we prove ΔBCD ~ΔCDE then we can say that /EDC =/BCD (by cpct)
so
BD =CE (given)
CD =CD (common)
1/2DE =BC (by mid point theorem)
ΔBCD ~ΔCDE
so by cpct /EDC =25°
(iii)/ADE +/EDC +/CDB =180° (linear pair)
/ADE +25°+100°=180°
/ADE +125°=180°
/ADE =180°-125°
/ADE =55°
(iv) /A +/D+/E =180°(sum of all angles of triangle)
55°+55°+/E =180°
110°+/E =180°
/E =180°-110°
/E =70° means
/AED =70°
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