Geometry Problem. Post your solution in the comment box below.

Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to view more details of problem 1221.

## Thursday, June 2, 2016

### Geometry Problem 1221: Intersecting Circles, Chord, Tangent, Parallel Chords, Collinear Points

Subscribe to:
Post Comments (Atom)

Since EF//CD => Arc(EC)=Arc(FD)

ReplyDeleteIn circle O , We have ∠ (DTB)= ½( Arc(EC)+Arc(DB))= ½(Arc(FD)+Arc(DB))= ½ Arc(FB)= ∠ (BAF)… (1)

Since DT tangent to circle Q at T => ∠ (BAT)= ∠ (DTB)…. (2)

Compare ( 1) to (2) we have ∠ (BAF)= ∠ (BAT)

So F, T , A are collinear

< DTB = < FET but < DTB = < TAB

ReplyDeleteSo < FET = < TAB

But < FET = < FAB

Hence < TAB = < FAB

So, F, T, A are collinear

Sumith Peiris

Moratuwa

Sri Lanka

Is arc EC=arc FD. Suppose that AT intersects the circle with center O to the point Κ,then <EKA=<EBA=<ATC=<KTD so EK//CD.Therefore arc KD=arc FD, so the points K, F will coincide.

ReplyDeleteTherefore points F,T,A are collinear.

Consider the line FA intersects the circle Q at P. Join AB.

ReplyDeleteP and T coincide and hence F,T,A are collinear

Ang EFA= Ang EBA= Ang CTA, because EF//CD => F, T, A collinear

ReplyDelete