Q.ABC is a triangle in which …

Given: triangle ABC is a right angled triangle, right angled at B, angle ABC = 90^o

           A circle is drawn, taking AB as diameter, intersects AC at P. PQ is a tangent, intersects at Q.

To prove: BQ = QC

Construction: Join BP.

Proof: PQ = BQ       ..........(i)       [Tangents drawn from an external point are equal]

          Therefore, angle PBQ = angle BPQ     [Angles opposite to equal side]

           Also,  angle APB = 90^o                        [Angle of semicircle]

           And angle APB + angle BPC = 180^o   [Linear pair]

           Now, angle BPC = 180^o - 90^o = 90^o

           Since, angle BPC + angle PBC + angle PCB = 180^o   [Angle sum property of a triangle]

           =>      angle PBC + angle PCB = 180^o - angle BPC = 180^{<font size="2">o</font>} - 90^{<font size="2">o </font>}= 90<font size="2">^o</font><font size="2">     .........(ii)</font>

          Now, angle BPC = 90^o

          =>        angle BPQ + angle CPQ = 90^o              ..........(iii)

          From eq. (i) and (ii), we get,

          angle PBC + angle PCB = angle BPQ + angle CPQ

          =>        angle PCQ = angle CPQ         [Since angle BPQ = angle PBQ]

          =>        PQ = QC             [In triangle PQC, opposite sides of equal angles]        ..........(iv)

          From, eq. (i) and (iv), we get,

           BQ = QC

          Hence proved.