--- In

Hyacinthos@yahoogroups.com, "ben_goss_ro" <ben_goss_ro@y...>

wrote:

> Here's a nice problem:

>

> Let the incircle of ABC touch CA and AB at E and H respectively.

Show

> that if H is on EF then HI passes through the midpoint of BC. (H is

> the orthocenter and I is the incenter)

Let BH,IF intersect at X; CH,IE intersect at Y

The figure IXHY enclosed by parallels BH,IE and CH,IF is a

parallelogram

So HI bisects XY...(i)

Angles IFE,IEF are equal, so are corresponding angles EHY,FHX

So triangles XHF, YHE are similar,

HX/HY = FH/EH = Sin FAH / Sin EAH = Cos B / CosC = HB / HC

Follows XY is parallel to BC ...(ii)

Hence from (i) & (ii)

HI bisects BC

Vijay