CHAPTER 9. GEOMETRY 9.2
Let∠FEA = y
∴∠FDA = y (∠’s subtended by same chord AF in cyclic quadrilateral FADE)
∴∠CBD = y (corresponding∠’s, FD� CB)
∴∠FEA =∠CBD
Step 3 : We have already proved 1 pair of angles equalin the previous question.
∠BCD =∠FAE (above)
Step 4 : Proving the last set ofangles equal is simply amatter of adding up theangles
in the triangles. Then we have proved similarity.
∠AFE = 180◦−x−y (∠’s in�AFE)
∠CDB = 180◦−x−y (∠’s in�CBD)
∴�AFE///�CBD (3∠’s equal)
Step 5 : This equation looks like it has to do with proportionality relation of similar
triangles. We already showed trianglesAFE andCBD similar in the previous question.
So lets start there.
DC
BD
=
FA
FE
∴
DC×FE
BD
= FA
Step 6 : Now we need to lookfor a hint about side FA. Looking at triangle CAH we
see that there is a line FG intersecting it parallelto base CH. This gives us another
proportionality relation.
AG
GH
=
FA
FC
(FG� CH splits up lines AH and AC proportionally)
∴ FA =
FC×AG
GH
Step 7 : We have 2 expressionsfor the side FA.
∴
FC.AG
GH
=
DC×FE
BD
