Question

PQ is a diameter of a circle with centre O. QR is a tangent to the circle equal to OQ. OT is perpendicular to PR. QT is produced to cut the circle at S. Can you prove that SP = ST?

Refer the image as a better option to reading.

Answer 1

Solution:

$\angle RQO = 90$

$OQ = QR$

So, $\angle QOR =\angle QRO$

$= 45$

$\left [ \angle QOR + \angle QRO = 90 \right ]$

$\angle OQR = 90$

OR is a diameter of a circle.

$\angle QOR =\angle QTR = 45$ [Angles of same segment]

$\angle QTR =\angle PTS = 45$   [Vertically opposite angles]

In $\triangle PST$

$\angle S + \angle P + \angle T = 180$

$90 + \angle P + 45 = 180$

Also, $\angle P = 45$

$\angle T = 45$

So, $\angle P = \angle T$

Thus, $SP = ST$.

Answer 2

Answer:

Step-by-step explanation:

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