Circle Geometry Question (1 Viewer)

[Nick_K]

Hi, I am stuck on a circle geometry problem and I'd appreciate any help I'm not sure how to put a picture of the diagram up here so I'll explain it as best I can.

Two circles with centres H and K touch at M (They are aligned left and right, with the circle centre H on the left and K on the right). A common tangent passes through M above the circles to a point R. Another line passing through R is tangent to both circles, touching them at points P and Q (P lies on the circle with centre H, Q lies on circle with centre K).

i) Show that quadrilaterals HPRM and MRQK are cyclic. (I have done this part, just thought I'd put it in to make is clearer.)

ii) Prove that triangles PRM and MKQ are similar.

Sorry if the description isn't clear, thank you for your time.

 

[Bored Of Fail 3]

thats easy

 

[Nick_K]

Could you explain it?

 

[benji_10]

∠PRM = ∠MKQ (int.∠ equal to opp. ext. angle of cyclic quads.)
PR = RM (tangents drawn to common ext. point (R) are equal in length)
therefore ΔPRM is isos.
MK = QK (radii egual in length)
therefore ΔMKQ is isos.

Hence, base angles of both triangles are equal as ∠PRM = ∠MKQ.
Hence, ΔMKQ///ΔPRM (equiangular)

 

[Nick_K]

Thank you.

 

[pwoh]

∠PRM = ∠MKQ (int.∠ equal to opp. ext. angle of cyclic quads.)
PR = RM (tangents drawn to common ext. point (R) are equal in length)
therefore ΔPRM is isos.
MK = QK (radii egual in length)
therefore ΔMKQ is isos.

Hence, base angles of both triangles are equal as ∠PRM = ∠MKQ.
Hence, ΔMKQ///ΔPRM (equiangular)

Excuse my ugly and possibly wrong diagram but how is ∠PRM the opposite interior angle to exterior ∠MKQ?

 

[Gussy Booo]

That is a terrible drawing
And he's talking about the external angles of a cyclic quadrilateral.
He is right

 

[pwoh]

Using my ugly diagram again, aren't these the external angles?

 

[Nick_K]

PRQ is a straight line, tangent to both circles. Sorry I explained that badly.

 

[pwoh]

PRQ is a straight line, tangent to both circles. Sorry I explained that badly.

Oh that makes a lot more sense, thanks