In triangle OBP and triangle ODQ, OB = OD, ∠POB = ∠QOD and alternate interior angles ∠OBP = ∠ODQ. By ASA congruence, triangle OBP ≅ triangle ODQ, proving OP = OQ; this direct triangle congruence is simplest. Cbse Class 9 Maths ...
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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 ...
Folding A onto B creates the perpendicular bisector crease across AB, marking its midpoint. Since this crease is parallel to side BC, by the Converse of the Midpoint Theorem it also bisects AC. Class 9 Ganita Manjari part 2 Chapter 12 ...
Diagonals of parallelogram ABCD bisect at O, so OA = OC and OB = OD. Since DP = BQ, subtracting gives OP = OQ. With diagonals AC and PQ bisecting each other, APCQ is a parallelogram. Class 9 Ganita Manjari Part ...
In triangles EAD and EBC, vertical angles at E are equal, so the four vertex angles sum to 360° – 2∠AED < 360°. Yes, make angle E nearly 179° so the remaining sum equals 2°. Cbse released Class 9 Ganita Manjari ...