## Monday, January 11, 2010

### Problem 417: Tangent circles, Tangent lines, Angles

Proposed Problem
Click the figure below to see the complete problem 417 about Tangent circles, Tangent lines, Angles.

Complete Problem 417
Level: High School, SAT Prep, College geometry

1. join E to A, D to B
draw DH perpendicular to EA

in ABDH
ang( HAB+ABD+BDH+DHA)= 164 + BDH + 90 (HAB+ABD=2∙82)
=> ang BDH = 106
=> ang HDF = 164

=> x = 16

2. solution needs to be improved
how did u get HAB + ABD = 2.82

3. thanks, I saw it today

ang ECD = 1/2 ( arc EC + arc CD ) (build common tang)

ang EAC + ang CBD = arc EC + arc CD

4. Problem 417
I design the common external tangent of two circles intersecting tinEF
in point K and DF intersects the KC at the point M. If <MCD=a=<MDC, then <KMF=2a.
But <KCE=82-2a=<KEC, so <FKC=184-2a.Then x+2a+184-2a=180 therefore x=16.
APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE

5. Extend FD to meet EC at G and let m(FEC) = x and m(GDC) = y

Join AE, AC and form the isosceles triangle AEC in which
m(AEC) = m(ECA) = 90-x

Similarly join CD, DB and form the isoceles trianlge CDB in which
m(CDB) = m(DCB) = 90-y

Since A,C and B are collinear we have
(90-x)+82+(90-y) = 180
=> x+y=82------------(1)

Now consider the triangle CGD. Since m(GCD) = 82 and m(GDC) = y, we have m(EGD) = 82+y
In the triangle EGF, we have
m(GEF) = x
m(EGF) = 82+y
Hence m(F) = 180-(x+82+y) = 180-(164) = 16 degrees (Substitute value of x+y from (1))

6. Let the common tangent at C meet EF at P and FD extended at Q.

PE = PC and DQ = QC so we realize that
< FPQ + < FQP = 2 X 82 = 164

Hence x = 180 - 164 = 16

Sumith Peiris
Moratuwa
Sri Lanka