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• ... Show ... 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)
Message 1 of 2 , May 2, 2004
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--- 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
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