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Question : 29 of 160
Marks:
+1,
-0
Solution:
Given,
A(a),B(b),C(c),D(d) are concyclic pointsand
xa+yb+zc+td=0 x+y+z+t=0 ‌=−‌
‌=‌ P=‌‌ or ‌P=‌ PA=‌−a=x(a−b) PB=‌−b=y(a−b) PC=‌−c=t(d−c) pD=‌−d=z(d−c) ∴‌‌PA×PB=PC×PD |x(a−b)||y(a−b)|=|t(d−c)|z(d−c)|
|xy||a−b|2=|zt|c−d|2
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