Exercise 1
If two lines intersect, prove that the vertically opposite angles are equal.
Not cross-checked
NCERT prints no numerical answer for this exercise, so this working has not been cross-checked against the book.
Let AB and CD be two lines intersecting at O.
\[\angle AOC + \angle AOD = 180^\circ \quad \text{(linear pair, ray OA on line CD)} \]
\[\angle AOD + \angle BOD = 180^\circ \quad \text{(linear pair, ray OD on line AB)} \]
\[\angle AOC + \angle AOD = \angle AOD + \angle BOD \]
\[\angle AOC = \angle BOD \]
Similarly, \(\displaystyle \angle AOD = \angle BOC\).
Answer: vertically opposite angles are equal.
\[\angle AOC + \angle AOD = 180^\circ \quad \text{(linear pair, ray OA on line CD)} \]
\[\angle AOD + \angle BOD = 180^\circ \quad \text{(linear pair, ray OD on line AB)} \]
\[\angle AOC + \angle AOD = \angle AOD + \angle BOD \]
\[\angle AOC = \angle BOD \]
Similarly, \(\displaystyle \angle AOD = \angle BOC\).
Answer: vertically opposite angles are equal.
\[\angle TBC = \tfrac{1}{2}\angle B \quad \text{(BT bisects } \angle B\text{)} \]
\[\angle ACD = 180^\circ-\angle C \quad \text{(linear pair)} \]
\[\angle TCD = \tfrac{1}{2}\angle ACD = 90^\circ-\tfrac{1}{2}\angle C \quad \text{(CT bisects } \angle ACD\text{)} \]
\[\angle BCT = 180^\circ-\angle TCD = 90^\circ+\tfrac{1}{2}\angle C \quad \text{(linear pair)} \]
\[\angle TBC+\angle BCT+\angle BTC = 180^\circ \quad \text{(angle sum, } \triangle BTC\text{)} \]
\[\angle BTC = 90^\circ-\tfrac{1}{2}(\angle B+\angle C) \]
\[\angle A+\angle B+\angle C=180^\circ \Rightarrow \angle B+\angle C = 180^\circ-\angle A \]
\[\angle BTC = 90^\circ - \tfrac{1}{2}(180^\circ-\angle A) = \tfrac{1}{2}\angle A \]
Answer: \(\displaystyle \angle BTC=\dfrac{1}{2}\angle BAC\).
\[\angle EPX = \angle EQY \quad \text{(corresponding angles, } WX\parallel ZY\text{)} \]
\[\angle EPR=\tfrac{1}{2}\angle EPX,\quad \angle EQS=\tfrac{1}{2}\angle EQY \quad \text{(PR, QS bisect them)} \]
\[\angle EPR=\angle EQS \]
\[PR \parallel QS \quad \text{(converse of corresponding angles axiom, transversal } EF\text{)} \]
Answer: the bisectors PR and QS are parallel.
\[\angle PMN=\angle PNM=90^\circ \quad \text{(both perpendicular to } AB\text{)} \]
\[\angle PMN+\angle PNM+\angle MPN=180^\circ \quad \text{(angle sum, } \triangle PMN\text{)} \]
\[90^\circ+90^\circ+\angle MPN=180^\circ \Rightarrow \angle MPN=0^\circ \]
This is impossible in a triangle, so M = N.
Answer: only one perpendicular can be drawn from P to AB.
\[EF\parallel GK,\ EF\perp AB \Rightarrow GK\perp AB \quad \text{(a line parallel to a perpendicular is itself perpendicular)} \]
\[GK\perp CD \quad \text{(given)} \]
\[GK\perp AB,\ GK\perp CD \Rightarrow AB\parallel CD \quad \text{(lines perpendicular to the same line are parallel)} \]
This contradicts AB and CD intersecting.
Answer: EF and GK must intersect.
\[\angle B+\angle C \geq 90^\circ+90^\circ = 180^\circ \]
\[\angle A+\angle B+\angle C=180^\circ \quad \text{(angle sum property)} \]
\[\angle A = 180^\circ-(\angle B+\angle C) \leq 0^\circ \]
This is impossible, since an angle of a triangle must be positive.
Answer: a triangle must have at least two acute angles.
\[\angle QPM = 90^\circ - \angle Q \quad \text{(angle sum, } \triangle PQM\text{, right angle at } M\text{)} \]
\[\angle QPR = 180^\circ - \angle Q - \angle R \quad \text{(angle sum, } \triangle PQR\text{)} \]
\[\angle QPA = \tfrac{1}{2}\angle QPR = 90^\circ - \tfrac{1}{2}(\angle Q + \angle R) \quad \text{(}PA\text{ bisects } \angle QPR\text{)} \]
\[\angle APM = \angle QPA - \angle QPM \quad \text{(ray } PM \text{ lies between } PQ \text{ and } PA\text{)} \]
\[\angle APM = \left[90^\circ - \tfrac{1}{2}(\angle Q+\angle R)\right] - \left[90^\circ - \angle Q\right] = \tfrac{1}{2}(\angle Q - \angle R) \]Answer: \(\displaystyle \angle APM = \dfrac{1}{2}(\angle Q - \angle R)\).