Q8 of 31 Page 218

Two circles have been drawn with centre A and B which touch each other externally at the point O. I draw a straight line passing through the point O intersects the two circles at P and Q respectively. Let us prove that AP || BQ


Capture.JPG


O is the point the two circles meet


AO and AP are the radius of circle A


BO and BQ are the radius of circle B


∠ POA = ∠ BOQ (Vertically Opposite)


∠ POA = ∠ OPA (Δ AOP is an isosceles triangle)


∠ BOQ = ∠ BQO (Δ BOQ is an isosceles triangle)


So combining the above two relations we can say


∠ OPA = ∠ BQO


Hence the above two angles forms a pair of alternate interior angle


So we say AP∥ BQ


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