Question

The bisectors of ㄥB and ㄥC of a ΔABC meet at O.
Show that ㄥBOC = 90° + ㄥA/2

Answer 1

ㄥA+ㄥB+ㄥC=180°
so,
ㄥA/2+ㄥB/2+ㄥC/2=90°
ㄥB/2+ ㄥC/2=90°- ㄥA/2........ equation (i)
now in ΔBOC,
ㄥOBC=ㄥB/2
ㄥOCB=ㄥC/2
ㄥOBC+ㄥOCB+ㄥBOC=180°
ㄥBOC=180°-(ㄥOBC+ㄥOCB)
ㄥBOC=180°-(ㄥB/2+ㄥC/2)
ㄥBOC=180°-(90°- ㄥA/2)....[from equation (i)]
ㄥBOC=180°-90°+ㄥA/2
ㄥBOC=90°+ㄥA/2