PQ and RS are two diameters of a circle with centre O Prove that the quadrilateral formed by the - Maths - Circles

Question

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

Akhil Goyal · Accepted Answer

Consider the
diagram below:
<img alt="" src=
"https://s3mn%20mnimgs%20com/img/shared/ck-files/ck_551029bb4d710%20png">
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, △APO≅△ARO (using SSS)Thus, ∠1=∠2 & ∠5=∠6Absolutely similarly, we can show that: △CSO≅△CQOTherefore we have:∠11=∠12 & ∠7=∠8Now, PQ and RS are straight lines, thus:∠5+∠6=∠8+∠7 (vertically opposite angles)which means that:2∠5=2∠7 (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=360⇒∠1+∠2+∠5+∠6=360-90-90=180Similarly, in □CQOS:∠11+∠12+∠7+∠8=180but, ∠7+∠8=∠5+∠6⇒∠1+∠2=∠11+∠12Since, ∠11=∠12 & ∠1=∠2∠2=∠12Since these are alternate angles, we have: AB∥CDAlso, since ∠1=∠11, we also have AD∥BDTherefore, 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

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.