(i) Diagonal BD meets EF at M. By Theorem 7, M and F bisect BD and BC. Adding segment lengths gives EF = (AB + CD)/2. (ii) Both methods work; using parallel line uniqueness is simpler. Class 9 Ganita Manjari Part ...
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(i) Diagonal BD meets EF at M. By Theorem 7, M and F bisect BD and BC. Adding segment lengths gives EF = (AB + CD)/2. (ii) Both methods work; using parallel line uniqueness is simpler. Class 9 Ganita Manjari Part ...