Q14 of 24 Page 157

ABCD is a squared figure. The two diagonals AC and BD intersect each other at the point O. The bisector of ∠BAC intersects BO at the point P and BC at the point Q. Let us prove that,


We have extended PQ till G such that OP||CG,


As O is intersection of Diagonals of a square, therefore O is mid point of AC.


Using the theorem,


The line segment joining the midpoints of two side of a triangle is parallel to the third side and equal to half of it, we get


……… (1)


As AP is angle bisector of ∠BAO,


⇒ ∠BAP = ∠OAP = θ


⇒ in ΔABQ, ∠AQB = 90 – θ


∠GQC = ∠AQB = 90 – θ (Opposite angles)


Similarly in ΔACG, as ∠ACG = 90° ,


⇒ ∠AGC = 90 – θ


As in ΔGQC, ∠AGC = ∠GQC = 90 – θ and as sides opposite to equal angles in a triangle are equal


⇒ CQ = CG ……… (2)


From equations (1) and (2),


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