Intersecting and Non-intersecting Lines Questions in English

(A) $XOY$ is a straight line.
Therefore,$\angle b + \angle a + \angle POY = 180^o$ (Linear pair axiom).
Given that $\angle POY = 90^o$,so $\angle b + \angle a + 90^o = 180^o$,which implies $\angle b + \angle a = 90^o$.
Given $a: b = 2: 3$,let $a = 2x$ and $b = 3x$.
Then $2x + 3x = 90^o$,so $5x = 90^o$,which gives $x = 18^o$.
Thus,$\angle a = 2 \times 18^o = 36^o$ and $\angle b = 3 \times 18^o = 54^o$.
Since lines $XY$ and $MN$ intersect at $O$,$\angle XON$ and $\angle MOY$ are vertically opposite angles,and $\angle XOM$ and $\angle NOY$ are vertically opposite angles.
Here,$\angle c = \angle XOM = \angle a + \angle POY$ is not correct based on the figure. Looking at the figure,$\angle c = \angle XON$.
Since $XOY$ is a straight line,$\angle XON + \angle NOY = 180^o$.
Also,$\angle NOY = \angle b = 54^o$ (vertically opposite angles).
Therefore,$\angle c = 180^o - 54^o = 126^o$.

(D) Let two lines $AB$ and $CD$ intersect at point $O$. Let $\angle AOC = 90^\circ$.
Since $AB$ is a straight line,$\angle AOC + \angle BOC = 180^\circ$ (Linear pair axiom).
$90^\circ + \angle BOC = 180^\circ \implies \angle BOC = 90^\circ$.
Similarly,$\angle AOC + \angle AOD = 180^\circ$ (Linear pair axiom).
$90^\circ + \angle AOD = 180^\circ \implies \angle AOD = 90^\circ$.
Finally,$\angle AOD + \angle BOD = 180^\circ$ (Linear pair axiom).
$90^\circ + \angle BOD = 180^\circ \implies \angle BOD = 90^\circ$.
Therefore,all other three angles are also right angles $(90^\circ)$.

(N/A) Let $l$ and $m$ be two intersecting lines. Let line $n$ be perpendicular to $l$ $(n \perp l)$ and line $p$ be perpendicular to $m$ $(p \perp m)$.
To prove: Lines $n$ and $p$ intersect each other.
Proof: Assume for the sake of contradiction that lines $n$ and $p$ are parallel to each other $(n \parallel p)$.
Since $n \perp l$ and $n \parallel p$,it follows that $p \perp l$ (because lines perpendicular to the same line are parallel,or conversely,a line perpendicular to one of two parallel lines is perpendicular to the other).
We are given that $p \perp m$. Thus,we have $p \perp l$ and $p \perp m$.
This implies that $l \parallel m$ (since both are perpendicular to the same line $p$).
However,this contradicts the given information that $l$ and $m$ are intersecting lines.
Therefore,our assumption that $n \parallel p$ must be false.
Hence,lines $n$ and $p$ must intersect each other.

(A) When two lines intersect,the vertically opposite angles are equal.
Since lines $AB$ and $CD$ intersect at $P$,$\angle APC$ and $\angle BPD$ are vertically opposite angles.
Therefore,$\angle APC = \angle BPD$.
Substituting the given expressions: $2x + 30^{\circ} = 4x - 20^{\circ}$.
Rearranging the terms: $30^{\circ} + 20^{\circ} = 4x - 2x$.
$50^{\circ} = 2x$,which gives $x = 25^{\circ}$.
Now,calculate the angles:
$\angle APC = 2(25) + 30 = 50 + 30 = 80^{\circ}$.
$\angle BPD = 4(25) - 20 = 100 - 20 = 80^{\circ}$.
Thus,$x = 25$,$\angle APC = 80^{\circ}$,and $\angle BPD = 80^{\circ}$.