ABCD is a cyclic quadrilateral whose diagonals AC and BD intersect at right angle at M. Prove that any line passing through M and bisecting any side of the quadrilateral is perpendicular to the opposite side.
Let ABCD be a cyclic quadrilateral whose diagonals AC and BD intersect at M at right-angle. MP, bisecting side CD when produced, meets AB in L.
To Prove: ∠PLB = 90°
Proof: ∠AMB = ∠BMC (given) ∠CMD = ∠DMA = 90°
DP = CP
∴ PM = PC = PD [...the median bisecting the hypotenuse of a right-triangle is half of the hypotenuse.]
Now in ΔCPM, MP = CP ⇒ ∠1 = ∠2
But ∠1 = ∠3 [Angles in the same segment of a circle]
and ∠2 = ∠4 (Vertically opposite angles)
... ∠3 = ∠4
But ∠4 + ∠5 = 90° (given)
⇒ ∠3 + ∠5 = 90°
Now in ΔMBL, ∠MBL + ∠BML + ∠MLB = 180°
⇒ ∠3 + ∠5 + ∠MLB = 180°
or 90° + ∠MLB = 180°
Hence, ∠MLB = 90° ⇒∠PLB = 90°
To Prove: ∠PLB = 90°
Proof: ∠AMB = ∠BMC (given) ∠CMD = ∠DMA = 90°
DP = CP
∴ PM = PC = PD [...the median bisecting the hypotenuse of a right-triangle is half of the hypotenuse.]
Now in ΔCPM, MP = CP ⇒ ∠1 = ∠2
But ∠1 = ∠3 [Angles in the same segment of a circle]
and ∠2 = ∠4 (Vertically opposite angles)
... ∠3 = ∠4
But ∠4 + ∠5 = 90° (given)
⇒ ∠3 + ∠5 = 90°
Now in ΔMBL, ∠MBL + ∠BML + ∠MLB = 180°
⇒ ∠3 + ∠5 + ∠MLB = 180°
or 90° + ∠MLB = 180°
Hence, ∠MLB = 90° ⇒∠PLB = 90°
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