From am external point $$P$$，tangents $$PA$$ and $$PB$$ are drawn to a circle. From a point $$Q$$ on the major (or minor) arc $$AB$$， perpendiculars are drawn to $$AB， PA$$，and $$PB$$. Prove that the perpendicular to $$AB$$ is the mean proportional between the other two perpendiculars.

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From am external point $$P$$，tangents $$PA$$ and $$PB$$ are drawn to a circle. From a point $$Q$$ on the major (or  minor) arc $$AB$$， perpendiculars are drawn to $$AB， PA$$，and $$PB$$. Prove that the perpendicular to $$AB$$ is the mean proportional between the other two perpendiculars.

Uniteasy Admin · Accepted Answer

Join point $$Q$$ to tangent points $$A$$ and $$B$$. 
According to Alternate Segment Theorem, the angle between a tangent and a chord is equal to the angle in the alternate segment.
Then,
$$\angle QBD = \angle QAE$$
So the right$$ 	riangle QBD$$ and $$	riangle QBD$$ are similar and the ratios of corresponding sides are equal.
$$\dfrac{QD}{QE}=\dfrac{QB}{QA} $$n
Similarly,
$$\angle QAC = \angle QBE$$
So the right$$ 	riangle QAC$$ and $$	riangle QBE$$ are similar and the ratios of corresponding sides are equal.
$$\dfrac{QC}{QE} = \dfrac{QA}{QB} $$n
Based on (1) and (2), we get 
$$\dfrac{QD}{QE} = \dfrac{QE}{QC} $$n
Therefore,
$$QE = \sqrt{QD\cdot QC} $$
Now we have proved that the perpendicular to $$AB$$ is the mean proportional between the other two perpendiculars

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