1. Let C be a point on a straight line AB. Circles are drawn with diameters AC and AB. Let P be any point on the circumference of the circle with diameter AB. If AP meets the other circle at Q, then
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By: anil on 05 May 2019 02.23 am
In $$ riangle$$ AQC, $$angle$$ AQC $$= 90^circ$$ ($$ecause$$ Angle in a semi circle is $$90^circ$$) and in $$ riangle$$ APB, $$angle$$ APB $$= 90^circ$$ ($$ecause$$ Angle in a semi circle is $$90^circ$$) Comparing two triangles $$ riangle$$ APB and $$ riangle$$ AQC, $$angle$$ QAC $$= angle PAB$$ $$angle$$ AQC $$= angle APB$$ $$ herefore riangle APB = riangle AQC$$ $$ herefore$$ QC // PB Since we cannot prove that C is exactly midpoint of AB, QC $$= frac{1}{2}$$PB cannot be proved
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