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Fig. 6.18A.Band AC of Δ A BC are produced to E andD. In Fig, 6.19, sidectiv. If angle bisectors BO and CO of 4CBE and ZBCDmeet each other at point O, then prove that;< BOC = 90°-4x2Fig. 6.19

Answer»

In triangle ABC, we apply angle sum property,therefore, x+y+z=180°=>y+x=180°-x. --------(1)Now, exterior angle EBC is sum of interior opposite angles x and z.so angle EBC= x+z=> angleEBC/2=(x+z)/2=>angleOBC=(x+z)/2. --------(2){OBC is half angle EBC as BO is angle bisector}

Similarly, angle DCB=x+yso angleDCB/2=(x+y)/2so angle OCB= (x+y)/2. --------(3)

now in triangle OBC we apply angle sum property.angle OBC + angle OCB + angle BOC=180°

using (2) and (3)(x+z)/2+(x+y)/2+angle BOC=180°=>x+(y+z)/2+angle BOC = 180°

Using (1)x+(180°-x)/2 + angle BOC= 180°=>x/2 +90° +angle BOC=180°=>angle BOC= 90°-x/2hence proved.


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