...and one more question.
ABCD is a cyclic quadrilateral. The edges AB, DC produced intersect at R. DA, CB produced intersect at S. Prove that the bisector of angles BRC and ASB are perpendicular.
Hey, arii.
Start by drawing up the missing sides
Below is a picture with some vital intersection points for ease of use.
\[ \text{Notice that } \angle ASO = \angle BSO \text{(SO is a bisector of } \angle ASB.\text{)} \\ \text{The same argument works with } \angle BRO \text{ and } \angle CRO.\]
\[ \text{Let } \angle ASO \text{ be } \theta \text{ and } \angle BRO \text{ be } \phi. \\ \text{Also, let } \angle ADR \text{ be } \gamma.\]
\[ \text{Now, using } \triangle SDC, \text{ we find that:} \\ \angle ASB + \angle SDC = \angle BCR \text{ (external angle of a triangle is equal to the sum of the interior angles)} \\ \text{So, } \angle BSR = 2\theta + \gamma. \]
\[ \text{Also, since ABCD is a cyclic quad, then: } \\ \angle SBA = \gamma \text{ (exterior angle of a cyclic quad is equal to the interior opposite angle). } \\ \therefore \angle CBR = \gamma \text{ (} \angle CBR = \angle SBA \text{ through vertically opp. angles)}\]
\[ 2\theta + 2\gamma + 2\phi = 180 \\ \theta + \gamma + \phi = 90. \]
\[ \text{Now notice that, in }\triangle QRC, \angle BCR + \angle CRQ + \angle CQR = 180 \text{ (angle sum of triangle)} \\ \angle BCR + \angle CRQ + \angle OQS = 180 \\ \angle OQS = 180 - (2\theta + \gamma + \phi).\]
\[ \text{Using } \triangle SOQ, \\ \angle SOQ + \angle OQS + \angle QSO = 180 \text{ (angle sum of triangle)} \\ \angle SOQ = 180 - 180 + (2\theta + \gamma + \phi) - \theta \\ \angle SOQ = 2\theta + \gamma + \phi - \theta = \theta + \gamma + \phi \\ \therefore \angle SOQ = 90^\circ\]
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