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math137

Member

Registered: 2014-08-07

Posts: 12

Geometric Proof

Hello,

How would you prove this:

Two circles are externally tangent at point P, as shown. Segment CPD is parallel to common external tangent AB. Prove that the distance between the midpoints of AB and CD is AB/2.

Thanks.

Last edited by math137 (2014-08-10 09:29:20)

Olinguito

Member

Registered: 2014-08-12

Posts: 649

Re: Geometric Proof

You have to do two things:

[list=1]
[*]Show that CD is twice as long as AB.[/*]
[*]Show that AC is perpendicular to BD[/*]
[/list]

1. is straightforward. Let E be the foot of the perpendicular from A to CP and F the foot of the perpendicular from B to PD. Then AEFB is a rectangle, AE bisects CP and BF bisects PD, and |CD| = |CP| + |PD| = |CE| + |EP| + |PF| + |FD| = 2|EP| + 2|PF| = 2|EF| = 2|AB|.

2. is more complicated. I’m afraid I can’t do it geometrically but I can do it using co-ordinate algebra. The working is too complicated so I’ll omit it here.

Using 1. and 2. we can solve the problem. Let Q be the midpoint of CD (so |CQ| = |QD| = |AB|). Then QB is perpendicular to BD since AC is perpendicular to BD and ACQB is a parallelogram. Thus QB is the perpendicular bisector in triangle QDG, where G is the point on BD extended on the side of B such that |DB| = |BG|. Also (by similar triangles) QG intersects AB at R, the midpoint of AB. Now triangle RBG is similar to triangle QDG and so is an isoscles triangle. Thus |RB| = |RG|. But by similar triangles |RG| = |RQ|. Thus |RB| = |RQ|. QED.

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Bassaricyon neblina

Enshrouded_

Member

Registered: 2015-07-31

Posts: 47

Re: Geometric Proof

How do you know that AE bisects CP and BF bisects PD?

evene

Member

Registered: 2015-10-18

Posts: 272

Re: Geometric Proof

Go to this website: http://www.mathisfunforum.com/viewtopic.php?id=22435

It has pretty much everything to the problem!

Last edited by evene (2015-11-29 12:25:59)