Extend PQ so QS = PQ. By SAS, △APQ ≅ △CSQ, making CS = AP = PB and CS ∥ PB. Thus PBCS is a parallelogram, giving PQ ∥ BC and PQ = BC/2. For (ii), complete PBCS similarly to ...
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No, they do not have to be midpoints. In a parallelogram ABCD, choosing points dividing opposite sides in equal ratios (like AP = CR = 1/3 AB and BQ = DS = 1/3 BC) creates an inscribed parallelogram that is ...
Yes, it holds in both cases. (i) The Midpoint Theorem on triangles formed by diagonals still ensures opposite sides are parallel and equal to half the diagonal. (ii) It degenerates into a segment when diagonal AC is parallel to BD ...
(i) MBND is a parallelogram, so MD ∥ BN. Applying Theorem 7 to △ABY and △CDX proves the intercepts along AC are equal. (ii) Construct a parallelogram with diagonal PQ, draw parallel lines dividing an auxiliary ray into three parts ...
Possibilities (iii), (iv) and (v) are possible. The sum of percentage shares must always equal 100%, so all shares cannot simultaneously increase or simultaneously decrease together, ruling out (i) and (ii). Class 9 Ganita Manjari Part 2 Chapter 10 Solutions Class ...