The bisectors of angleB and angleC of quadrilateral ABCD meet at a poi

1.	The bisectors of angleB and angleC of quadrilateral ABCD meet at a point R inside the quadrilateral. If angleA=95 and angleD=111 , find angleBRC
Answer» Let the quadrilateral ABCD have the angles <A = 2a, <B = 2b, <C = 2c and <D = 2d. The bisectors of <B and <C meet at R. So <BRC = 180 -b-c. The sum of the four angles <A+<B+<C+<D + 360 deg = 2a+2b+2c+2d, or a+b+c+d = 180 deg …(1) we have seen that <BRC = 180-b-c = 180 -(b+c) …(2). Put the value of b+c from (1) in (2) to get <BRC = 180 -[180-a-d] = 180 -180+a+d, or <BRC = a+d = (1/2)[<A+<D]. =1/2(95+111)=1/2(206)=103°

The bisectors of angleB and angleC of quadrilateral ABCD meet at a point R inside the quadrilateral. If angleA=95 and angleD=111 , find angleBRC

Answer»

Let the quadrilateral ABCD have the angles <A = 2a, <B = 2b, <C = 2c and <D = 2d.

The bisectors of <B and <C meet at R. So <BRC = 180 -b-c.

The sum of the four angles <A+<B+<C+<D + 360 deg = 2a+2b+2c+2d, or

a+b+c+d = 180 deg …(1)

we have seen that <BRC = 180-b-c = 180 -(b+c) …(2). Put the value of b+c from (1) in (2) to get

<BRC = 180 -[180-a-d] = 180 -180+a+d, or

<BRC = a+d = (1/2)[<A+<D]. =1/2(95+111)=1/2(206)=103°