Questionathematics/P2 13 FET - Grade 11 MDE Novembe In the diagram, PQ is a tangent at Q. PRS is a secant od circle RSTWQ. RW cuts S QT at L. PS || QT and RS = TW. R 3 2 ove that: .2.1 KQ is a tangent to circle LQW. 2.2 PRQ=RIQ 2.3 2.4 2.5 RÎQ=KOP PRKQ is a cyclic quadrilateral. RSLQ is not a cyclic quadrilateral. 4 S 2 1 K 2 W GRAND TOTA
Studdy Solution
STEP 1
Assumptions
1. PQ is a tangent to the circle at point Q.
2. PRS is a secant line intersecting the circle at points R and S.
3. RW is a line intersecting QT at point L.
4. Lines PS and QT are parallel.
5. RS is equal to TW.
6. We need to prove several geometric properties about the given diagram.
STEP 2
Prove that KQ is a tangent to circle LQW.
Since PQ is a tangent to the circle at Q, by the tangent-secant theorem, the angle between the tangent and the chord through the point of contact is equal to the angle in the alternate segment. Therefore, we need to show that angle KQL is equal to angle KWL.
STEP 3
Prove that PRQ = RIQ.
Since PS || QT, by the alternate interior angles theorem, angle PRQ is equal to angle RIQ because they are alternate interior angles formed by the transversal PR.
STEP 4
Prove that RÎQ = KOP.
Since PS || QT and RS = TW, triangles PRS and PQT are similar by the AA criterion (angle-angle similarity). Therefore, angle RÎQ is equal to angle KOP as corresponding angles of similar triangles.
STEP 5
Prove that PRKQ is a cyclic quadrilateral.
A quadrilateral is cyclic if the opposite angles are supplementary. We need to show that angle PRK + angle PQK = 180°.
STEP 6
Prove that RSLQ is not a cyclic quadrilateral.
A quadrilateral is not cyclic if the opposite angles are not supplementary. We need to show that angle RSL + angle RQL ≠ 180°.
STEP 7
Conclusion
By proving each of the above steps, we have demonstrated the required geometric properties of the diagram.
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