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तमिलनाडु बोर्ड ऑफ सेकेंडरी एज्युकेशनएसएसएलसी (अंग्रेजी माध्यम) कक्षा १०

O is any point inside a triangle ABC. The bisector of ∠AOB, ∠BOC and ∠COA meet the sides AB, BC and CA in point D, E and F respectively. Show that AD × BE × CF = DB × EC × FA - Mathematics

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प्रश्न

O is any point inside a triangle ABC. The bisector of ∠AOB, ∠BOC and ∠COA meet the sides AB, BC and CA in point D, E and F respectively. Show that AD × BE × CF = DB × EC × FA

योग

उत्तर

In ∆ABC the bisector meets AB at D, BC at E and AC at F.

The angle bisector AO, BO and CO intersect at “O”.

By Cevas Theorem

`"AD"/"DB" xx "BF"/"EC" xx "CF"/"AF"` = 1

AD × BE × CF = DB × EC × AF

Hence it is proved

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अध्याय 4: Geometry - Unit Exercise – 4 [पृष्ठ २००]

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सामाचीर कलवी Mathematics [English] Class 10 SSLC TN Board
अध्याय 4 Geometry
Unit Exercise – 4 | Q 3 | पृष्ठ २००
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