Answer :

Given:

Non-parallel sides PS and QR of a trapezium PQRS are equal.

Construction:

Draw \(SM\perp{PQ}\) and \(RN\perp{PQ}\).

To prove: ABCD is cyclic trapezium.

Proof:

In \(\triangle{SMP}\) and \(\triangle{RNQ}\), we have,

SP = RQ ...(Given)

SM = RN

(Distance between two parallel twolines is always equal)

and \(\angle{SMP}\) = \(\angle{RNQ}\)

(Each \(90^\circ\))

By RHS criterion, we get,

\(\triangle{SMP}\) \(\displaystyle \cong\) \(\triangle{RNQ}\)

Hence, \(\angle{P}\) = \(\angle{Q}\) ...(By CPCT)

Also, \(\angle{PSM}\) = \(\angle{QRN}\)

Now, adding \(90^\circ\) on both sides, we get,

\(\therefore \) \(\angle{PSM}\) + \(90^\circ\) = \(\angle{QRN}\) + \(90^\circ\)

\(\therefore \) \(\angle{MSR}\) + \(\angle{PSM}\) = \(\angle{NRS}\) + \(\angle{QRN}\)

(Since, \(\angle{MSR}\) = \(\angle{NRS}\) = \(90^\circ\))

\(\therefore \) \(\angle{PSR}\) = \(\angle{QRS}\)

\(\Rightarrow \) \(\angle{S}\) = \(\angle{R}\)

\(\therefore \) \(\angle{P}\) = \(\angle{Q}\) and \(\angle{S}\) = \(\angle{R}\)

\(\because \) Sum of the angles of a quadrilateral is \(360^\circ\)

We get,

\(\angle{P}\) + \(\angle{Q}\) + \(\angle{S}\) + \(\angle{R}\) = \(360^\circ\)

From eq. (i), we get,

2[\(\angle{Q}\) + \(\angle{S}\)] = \(360^\circ\)

\(\therefore \) [\(\angle{Q}\) + \(\angle{S}\)] = \(180^\circ\)

Hence, it is proved that PQRS is a cyclic trapezium.

- In figure A, B and C are three points on a circle with centre O such that \(\angle{BOC}\) = \(30^\circ\) and \(\angle{AOB}\) = \(60^\circ\). If D is a point on the circle other than the arc ABC, find \(\angle{ADC}\).Let PQ = QR = PR = X
- A chord of a circle is equal to the radius of the circle. Find the angle subtended by the chord at a point on the minor arc and also at a point on the major arc.
- In figure, \(\angle{PQR}\) = \(100^\circ\) ,where P, Q and R are points on a circle with centre O. Find \(\angle{OPR}\).
- In figure, \(\angle{ABC}\) = \(69^\circ\), \(\angle{ACB}\) = \(31^\circ\). Find \(\angle{BDC}\).
- In figure, A, B and C are four points on a circle. AC and BD intersect at a point E such that \(\angle{BEC}\) = \(130^\circ\) and \(\angle{ECD}\) = \(20^\circ\). Find \(\angle{BAC}\).
- ABCD is a cyclic quadrilateral whose diagonals intersect at a point E. If \(\angle{DBC}\) = \(70^\circ\), \(\angle{BAC}\) = \(30^\circ\), Find \(\angle{BCD}\). Further, if AB = BC, find \(\angle{EDC}\).
- If diagonals of a cyclic quadrilateral are diameters of the circle through the vertices of the quadrilateral, prove that it is a rectangle.
- Two circles intersect at two points B and C. Through B, two line segments ABD and PBQ are drawn to intersect the circles at A, D and P, Q respectively (see figure). Prove that \(\angle{ACP}\) = \(\angle{QCD}\).
- If circles are drawn taking two sides of a triangle as diameters, prove that the point of intersection of these circles lie on the third side.
- ABC and ADC are two right angled triangles with common hypotenuse AC. Prove that \(\angle{CAD}\) = \(\angle{CBD}\).
- Prove that a cyclic parallelogram is a rectangle.

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