enter image description herethere is a line AD whose midpoint is O (AO=OD). a semicircle is drawn with centre O and any radius < AO with its straight edge being part of line AD. lines AB and CD are drawn tangent to the semicircle's arc such that BC is also a tangent to the arc. the question is to prove that $AO^2=AB·CD$

  • $\begingroup$ how do I paste an image in? $\endgroup$ – stanley dodds Mar 27 '15 at 7:45
  • $\begingroup$ click on edit. You will find a picture icon in the window, using that upload your pic. $\endgroup$ – MonK Mar 27 '15 at 9:06

I have made minor changes with suitable labels making presentation easier to read. (See below.)

enter image description here

Adopting the result of the special case shown earlier, we have $OA^2 = AB’. XD$.

i. e. $AO^2 = x.(t + z + x) = xt + xz + x^2$ … (1)

The target is to have $AO^2 = AB.CD$

i. e. $AO^2 = (x + y).(z + x) = xz + yz + x^2 + xy $ … (2)

Comparing the two, our target becomes "is $xt = y(x + z)$"

i. e. $\frac {x}{y} = \frac {x + z}{t}$

i. e. $\frac {x + y}{y} = \frac {x + z + t}{t}$

Construction:- (1) Join B’C’. (2) Join X O. (3) Through C, draw CE // DA cutting XO at G.

Now, by similar triangles, $\frac {x + z + t}{t} = \frac {XO}{XG}$.

There are many triangles that are similar to ⊿BAH and ⊿BB’ K. Examples are ⊿XEG and ⊿XAO.

From them, I am quite sure that $\frac {x + y}{y} = … = \frac {XO}{XG}$

That completes the proof.

Remark: I am also sure that there are more elegant ways of proving this fact.


enter image description here

I hope my figure was drawn correctly according to your question. In addition, I have added the red and green circles to show the exact locations of B and T, the points of contacts of the tangents AC and DC.

It is quite obvious that the objects are symmetric about the y-axis.

$AO^2 = DO^2$

$= CD^2 – CO^2$ … [Pythagoras theorem]

$= CD^2 – CD.CT$ … [power of a point]

$= CD.(CD – CT)$

$= CD.(CA - CB)$

$= CD.AB$

  • $\begingroup$ thank you for the answer, but unfortunately this is not exactly what the question's diagram shows, although it is one special case of the given conditions. line BC is not necessarily an extention of line AB, in almost every case it is at an angle to AB, therefore also most of the time B is not the point where AB (or BC) is tangent to the semicircle. as in the title, ABCD forms a quadrilateral, with AB, BC and CD tangents to a circle and AD crossing through the centre O which is the midpoint. $\endgroup$ – stanley dodds Mar 27 '15 at 6:56
  • $\begingroup$ @stanleydodds Without a diagram, it is quite difficult to guess the description exactly. I think now I get what you mean. However, with the above suggested solution is still true (for the special case). It might still be of some value for the workout of the required solution. It would be quite interesting if we work things backward. I guess the problem is now:- (1) B’ is a point on and between B and C. (2) C’ is a point on and between C and T. (3) B’C’ is tangent to the circle at N. The target question now becomes proving $AO^2 = AB’.C’D$ (given that $AO^2 = AB.CD$ is true). $\endgroup$ – Mick Mar 27 '15 at 9:07
  • $\begingroup$ ok yes. the problem is that I am using a phone and cannot paste images. I have used geogebra to create a diagram but cannot put it in here. $\endgroup$ – stanley dodds Mar 27 '15 at 15:35
  • $\begingroup$ I have added an image, hopefully it is clearer now. $\endgroup$ – stanley dodds Mar 27 '15 at 15:47

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