﻿ From am external point P，tangents PA and PB are drawn to a circle. From a

#### Question

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.

Collected in the board: Circles problems

Steven Zheng posted 1 day ago

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 \triangle QBD and \triangle QBD are similar and the ratios of corresponding sides are equal.

\dfrac{QD}{QE}=\dfrac{QB}{QA}
(1)

Similarly,

\angle QAC = \angle QBE

So the right \triangle QAC and \triangle QBE are similar and the ratios of corresponding sides are equal.

\dfrac{QC}{QE} = \dfrac{QA}{QB}
(2)

Based on (1) and (2), we get

\dfrac{QD}{QE} = \dfrac{QE}{QC}
(3)

Therefore,

QE = \sqrt{QD\cdot QC}

Now we have proved that the perpendicular to AB is the mean proportional between the other two perpendiculars

Steven Zheng posted 1 day ago

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