In the diagram $\mathrm{O}$O is the center of the circle. $\mathrm{PS}$ is the tangent of the circle at $\mathrm{S}$. $\angle \mathrm{OSQ}$$= 50°$. Find $\angle \mathrm{SPR}.$

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In the given diagram $\mathrm{ABCD}$is a trapezium. If $\mathrm{AB} = 4 \mathrm{cm}, \mathrm{CD} = 6 \mathrm{cm} \mathrm{and} \mathrm{OB} = 5 \mathrm{cm}, \mathrm{then} \mathrm{BD} =\mathrm{\_\_\_\_\_cm}.$

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In a right angle triangle $\mathrm{BAC}$given below, $\text{AD }$is the altitude of the hypotenuse $\mathrm{BC}$. The figure is followed by three possible inferences.

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$1$. Triangle $\mathrm{ABD}$and triangle $\mathrm{CAD}$are similar

$2$. Triangle $\mathrm{ADB}$and triangle $\mathrm{CDA}$are congruent.

$3$. Triangle $\mathrm{ADB}$and triangle $\mathrm{CAB}$are similar.