(i) By Varignon’s Theorem, PQRS is a parallelogram, so its diagonals PR and QS bisect each other. (ii) PQ = RS = AC/2 and QR = SP = BD/2. If AC = BD, PQRS is a rhombus, having perpendicular diagonals. Yes, ...
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Let K be the midpoint of AD. By the Midpoint Theorem, KH ∥ AB and KG ∥ CD. Since AB ∥ CD, K, H, G are collinear, making GH ∥ AB. If AB = CD, ABCD is a parallelogram where ...
By the Midpoint Theorem in △ABC, MN ∥ BC, so MN ∥ BD. In △ABD, line MN passes through the midpoint M of AB parallel to BD, so by Theorem 7 it bisects AD. Cbse Class 9 Maths Ganita Manjari Part ...
2587392 mod 157656 = 64896, reducing to gcd(157656, 64896). Then 157656 mod 64896 = 27864, 64896 mod 27864 = 9168, 27864 mod 9168 = 360, 9168 mod 360 = 168, 360 mod 168 = 24 and 168 mod 24 = ...
1789287 mod 237656 = 125695, reducing to gcd(237656, 125695). Then 237656 mod 125695 = 111961, then remainder 13734, then 2089, then 1200, 889, 311, 267, 44, 3, 2 and finally remainder 1 followed by 0, yielding GCD 1. Cbse Class 9 ...