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Question 1: The side QR of $\triangle PQR$ is produced to a point S. If the bisectors of $\angle PQR$ and $\angle PRS$ meet at point T, then prove that $\angle QTR = \frac{1}{2} \angle QPR$.
Solution: In $\triangle TQR$, ext $\angle TRS = \angle TQR + \angle QTR$. In $\triangle PQR$, ext $\angle PRS = \angle PQR + \angle QPR$.
Since $2\angle TRS = \angle PRS$ and $2\angle TQR = \angle PQR$, substitute and solve to get $\angle QTR = \frac{1}{2} \angle QPR$. -
Question 2: In $\triangle ABC$, bisectors BO and CO of exterior angles $\angle CBE$ and $\angle BCD$ meet at O. Prove that $\angle BOC = 90^\circ - \frac{1}{2} \angle A$.
Solution: $\angle OBC = \frac{1}{2}(180 - B)$, $\angle OCB = \frac{1}{2}(180 - C)$. In $\triangle OBC$, $\angle BOC = 180 - (\angle OBC + \angle OCB)$. Simplify using $A+B+C=180$.
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Question 3: If two parallel lines are intersected by a transversal, prove that the bisectors of the two pairs of interior angles enclose a rectangle.
Solution: Bisectors of consecutive interior angles meet at $90^\circ$ (since sum is $180^\circ$, half sum is $90^\circ$). Since all 4 angles are $90^\circ$, it is a rectangle.
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Question 4: POQ is a line. Ray OR is perpendicular to line PQ. OS is another ray lying between rays OP and OR. Prove that $\angle ROS = \frac{1}{2}(\angle QOS - \angle POS)$.
Solution: $\angle ROS = \angle QOS - \angle QOR = \angle QOS - 90^\circ$. Also $\angle ROS = 90^\circ - \angle POS$. Add both equations: $2\angle ROS = \angle QOS - \angle POS$.
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Question 5: In a triangle ABC, the bisector of $\angle A$ meets BC at D. Prove that $\angle ADB > \angle C$.
Solution: In $\triangle ADC$, $\angle ADB$ is the exterior angle. So $\angle ADB = \angle DAC + \angle C$. Since $\angle DAC > 0$, $\angle ADB > \angle C$.
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Question 6: Two lines are respectively perpendicular to two parallel lines. Show that they are parallel to each other.
Solution: Let $l || m$. $p \perp l$ and $q \perp m$. Since $p \perp l$ and $l || m$, $p \perp m$. Now $p \perp m$ and $q \perp m$, so $p || q$ (lines perpendicular to same line are parallel).
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Question 7: Find the angle which is four times its complement.
Solution: Let angle be $x$. Complement is $90-x$.
$x = 4(90-x) \Rightarrow x = 360 - 4x \Rightarrow 5x = 360 \Rightarrow x = 72^\circ$. -
Question 8: If the bisectors of angles $\angle B$ and $\angle C$ of a triangle $ABC$ meet at a point $O$, then prove that $\angle BOC = 90^\circ + \frac{1}{2} \angle A$.
Solution: In $\triangle OBC$, $\angle BOC = 180 - \frac{1}{2}(B+C)$. Since $B+C = 180-A$, $\angle BOC = 180 - \frac{1}{2}(180-A) = 180 - 90 + A/2 = 90 + A/2$.
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Question 9: In Fig, $AB || CD$ and $CD || EF$. Also $EA \perp AB$. If $\angle BEF = 55^\circ$, find the values of $x, y$ and $z$.
Solution: (Assuming standard figure) $y + 55 = 180$ (consecutive interior) $\Rightarrow y = 125$. $x = y$ (corresponding) $\Rightarrow x = 125$. $z = 90 - 55 = 35$ (if Z is inside).
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Question 10: Prove that if the arms of one angle are respectively parallel to the arms of another angle, then the two angles are either equal or supplementary.
Solution: Extend arms to intersect. Use corresponding angles property twice. If both are acute/obtuse, they are equal. If one acute one obtuse, they are supplementary.