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https://photos.app.goo.gl/v7KgdZECAesRBJjr9Draw altitude BD of triangle ABC ( see sketch)We have B, H, D are collinear ( H is the orthocenter)Quadrilateral ADEB is cyclic so HE.HA=HB.HD= 140In cycle BTDC we have HT.HF=HB.HD=140So HF= 140/HT=8.5
minor correction of the last line:So HF= 140/HT=8.75
Draw altitude BPWe have BH . HP = 140=> TH . HF = 140 => HF = 8.75
https://photos.app.goo.gl/7F46Z7YkTbVG5kkk9
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https://photos.app.goo.gl/v7KgdZECAesRBJjr9
ReplyDeleteDraw altitude BD of triangle ABC ( see sketch)
We have B, H, D are collinear ( H is the orthocenter)
Quadrilateral ADEB is cyclic so HE.HA=HB.HD= 140
In cycle BTDC we have HT.HF=HB.HD=140
So HF= 140/HT=8.5
minor correction of the last line:
DeleteSo HF= 140/HT=8.75
Draw altitude BP
ReplyDeleteWe have BH . HP = 140
=> TH . HF = 140 => HF = 8.75
https://photos.app.goo.gl/7F46Z7YkTbVG5kkk9
Delete