perplexus.info :: Geometry : Sum of Two Angles

Sum of Two Angles (Posted on 2010-08-01)

Let D and E be points on sides AB and AC respectively of ΔABC.
Let the bisectors of /ABE and /ACD intersect in point F.

Prove that /BDC + /BEC = 2 /BFC.

Submitted by Bractals

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Solution: (Hide)

From ΔACD, /BDC = /BAC + /ACF + /DCF (1) From ΔABE, /BEC = /BAC + /ABF + /EBF (2) Adding equations (1) and (2), /BDC + /BEC = 2 /BAC + (/ACF + /DCF) + (/ABF + /EBF) = 2 (/BAC + /ACF + /ABF) (3) From ΔABC and ΔBCF, /BAC + /CBE + /ABF + /EBF + /BCD + /ACF + /DCF = 180° = /BFC + /CBE + /EBF + /BCD + /DCF or /BFC = /BAC + /ACF + /ABF (4) Combining equations (3) and (4), /BDC + /BEC = 2 /BFC

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	Subject	Author	Date
	Solution	Harry	2010-08-03 20:16:17

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