PQ and RS are two diameters of a circle with centre O. Prove that the quadrilateral formed by the tangents at the extremities of these diameters is a parallelogram. Also show that the diagonals of this parallelogram pass through O i.e. the centre.

Consider the diagram below:

We need to prove that ABCD is a parallelogram. To prove this, we will first prove that line AC and BD passes through O.
First, join AO and consider APO and AROWe know that AP=AR (tangents drawn from the same external point are of same length)PO=RO=rand, AO is commonSo, APOARO (using SSS)Thus, 1=2 & 5=6Absolutely similarly, we can show that: CSOCQOTherefore we have:11=12 & 7=8Now, PQ and RS are straight lines, thus:5+6=8+7 (vertically opposite angles)which means that:25=27 (since 5=6 & 7=8)5=7Similarly, 6=8This means that A,O,C must be collinear (since the vertically opposite angles are equal). Therefore AC passes through OBy analogy and symmetry, we can say that BD will also pass through ONow consider APOR:1+2+3+4+5+6=3601+2+5+6=360-90-90=180Similarly, in CQOS:11+12+7+8=180but, 7+8=5+61+2=11+12Since, 11=12 & 1=22=12Since these are alternate angles, we have: ABCDAlso, since 1=11, we also have ADBDTherefore, in ABCD both the pairs of opposite sides are parallel and thus, ABCD is a parallelogramSecond part of the question is already proved above, ie, both the diagonals pass through O & thus, intersect at O.

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