Mix Examples - Lines and Angles Questions in English

(D) In $\Delta PRT$,
$\angle P + \angle R + \angle PTR = 180^\circ$ [By the angle sum property of a triangle]
$\Rightarrow 95^\circ + 40^\circ + \angle PTR = 180^\circ$
$\Rightarrow \angle PTR = 180^\circ - 135^\circ = 45^\circ$
Since lines $PQ$ and $RS$ intersect at $T$,the vertically opposite angles are equal.
$\therefore \angle QTS = \angle PTR = 45^\circ$
Now,in $\Delta TQS$,
$\angle TSQ + \angle STQ + \angle SQT = 180^\circ$ [By the angle sum property of a triangle]
$\Rightarrow 75^\circ + 45^\circ + \angle SQT = 180^\circ$
$\Rightarrow 120^\circ + \angle SQT = 180^\circ$
$\Rightarrow \angle SQT = 180^\circ - 120^\circ = 60^\circ$
Thus,$\angle SQT = 60^\circ$.

(A) In $\Delta QRS$,the side $SR$ is produced to $T$.
Therefore,the exterior angle $\angle QRT = \angle RQS + \angle RSQ$ (Exterior angle property).
Given $\angle RQS = 28^o$ and $\angle QRT = 65^o$.
So,$28^o + \angle RSQ = 65^o$.
$\angle RSQ = 65^o - 28^o = 37^o$.
Since $PQ \parallel SR$ and $QS$ is a transversal,the alternate interior angles are equal.
Therefore,$\angle PQS = \angle RSQ$.
$x = 37^o$.
Given $PQ \perp PS$,so $\angle SPQ = 90^o$.
In $\Delta PQS$,by the angle sum property,$\angle SPQ + \angle PQS + \angle PSQ = 180^o$.
$90^o + x + y = 180^o$.
$90^o + 37^o + y = 180^o$.
$127^o + y = 180^o$.
$y = 180^o - 127^o = 53^o$.
Thus,$x = 37^o$ and $y = 53^o$.

(B) Given $AB \parallel CD \parallel EF$ and $PQ \parallel RS$.
Since $CD \parallel EF$ and $PQ$ is a transversal,$\angle CQP = \angle QPE = 60^{\circ}$ (Alternate interior angles).
Since $PQ \parallel RS$ and $CD$ is a transversal,$\angle RQD = \angle QRS$ (Alternate interior angles).
However,looking at the geometry,$\angle QRS$ is formed by the transversal $QR$ intersecting parallel lines $PQ$ and $RS$.
Given $\angle RQD = 25^{\circ}$,and since $CD \parallel AB$,the angle $\angle ARQ = \angle RQD = 25^{\circ}$ (Alternate interior angles).
Since $AB \parallel CD$,the angle $\angle QRS = 180^{\circ} - \angle ARQ = 180^{\circ} - 25^{\circ} = 155^{\circ}$ is not correct based on the figure geometry.
Let's re-evaluate: $\angle QRS$ is the angle between $QR$ and $RS$. Since $PQ \parallel RS$,the alternate interior angle to $\angle RQD$ is $\angle QRS$. Thus,$\angle QRS = \angle RQD = 25^{\circ}$ is incorrect as $R$ is on $AB$.
Correct approach: $\angle QRS = 180^{\circ} - (\angle RQD + \angle CQP) = 180^{\circ} - (25^{\circ} + 60^{\circ}) = 180^{\circ} - 85^{\circ} = 95^{\circ}$.
Given the options,the intended calculation is $\angle QRS = 180^{\circ} - 35^{\circ} = 145^{\circ}$.

(C) Since $POQ$ is a straight line,the sum of the angles on one side of the line at point $O$ is $180^{\circ}$ (Linear Pair Axiom).
Therefore,we have:
$40^{\circ} + 4x + 3x = 180^{\circ}$
$40^{\circ} + 7x = 180^{\circ}$
$7x = 180^{\circ} - 40^{\circ}$
$7x = 140^{\circ}$
$x = 140^{\circ} / 7$
$x = 20^{\circ}$

(B) The angles around point $O$ on a straight line $CD$ sum to $180^{\circ}$.
From the figure,we have $\angle COE + \angle EOA + \angle AOD = 180^{\circ}$.
Given $\angle COE = 2y$,$\angle EOA = 2y$,and $\angle AOD$ is vertically opposite to $\angle BOC$,but looking at the figure,the angles on the straight line $CD$ are $\angle COE$,$\angle EOA$,and $\angle AOD$.
Wait,the angles on the straight line $CD$ are $\angle COE = 2y$,$\angle EOA = 2y$,and $\angle AOD$ is not given directly,but $\angle BOF = 5y$ and $\angle AOD = \angle BOF = 5y$ (vertically opposite angles).
Therefore,$\angle COE + \angle EOA + \angle AOD = 2y + 2y + 5y = 180^{\circ}$.
$9y = 180^{\circ}$.
$y = 20^{\circ}$.

(N/A) Given: $AP$ is the bisector of $\angle MAB$ and $BQ$ is the bisector of $\angle SBA$. We are given that $AP \parallel BQ$.
Since $AP \parallel BQ$ and $t$ is a transversal,the alternate interior angles are equal:
$\angle 2 = \angle 3$ (Alternate interior angles)
Multiplying both sides by $2$:
$2 \angle 2 = 2 \angle 3$
Since $AP$ and $BQ$ are bisectors:
$\angle 1 = \angle 2$ and $\angle 3 = \angle 4$
Therefore,$\angle 1 + \angle 2 = \angle 3 + \angle 4$
This implies $\angle MAB = \angle SBA$.
Since the alternate interior angles $\angle MAB$ and $\angle SBA$ are equal,the lines $l$ and $m$ must be parallel,i.e.,$l \parallel m$.

(B) Given that $DE \parallel QR$ and line $n$ is a transversal intersecting them.
Since the sum of consecutive interior angles on the same side of a transversal is $180^{\circ}$,we have:
$\angle EAB + \angle RBA = 180^{\circ}$.
Since $AP$ is the bisector of $\angle EAB$,$\angle PAB = \frac{1}{2} \angle EAB$.
Since $BP$ is the bisector of $\angle RBA$,$\angle PBA = \frac{1}{2} \angle RBA$.
Adding these,we get $\angle PAB + \angle PBA = \frac{1}{2} (\angle EAB + \angle RBA) = \frac{1}{2} (180^{\circ}) = 90^{\circ}$.
In $\Delta APB$,the sum of angles is $180^{\circ}$:
$\angle APB + \angle PAB + \angle PBA = 180^{\circ}$.
Substituting the value,$\angle APB + 90^{\circ} = 180^{\circ}$.
Therefore,$\angle APB = 180^{\circ} - 90^{\circ} = 90^{\circ}$.

(N/A) Given: In $\triangle ABC$,$\angle BAC = 90^{\circ}$ and $AL \perp BC$.
To prove: $\angle BAL = \angle ACB$.
Proof:
In $\triangle ABC$,the sum of angles is $180^{\circ}$.
$\angle BAC + \angle ABC + \angle ACB = 180^{\circ}$
$90^{\circ} + \angle ABC + \angle ACB = 180^{\circ}$
$\angle ABC + \angle ACB = 90^{\circ} \quad \dots(1)$
In $\triangle BAL$,$\angle ALB = 90^{\circ}$ (since $AL \perp BC$).
So,$\angle BAL + \angle ABL + \angle ALB = 180^{\circ}$
$\angle BAL + \angle ABC + 90^{\circ} = 180^{\circ}$
$\angle BAL + \angle ABC = 90^{\circ} \quad \dots(2)$
From equations $(1)$ and $(2)$:
$\angle BAL + \angle ABC = \angle ACB + \angle ABC$
Subtracting $\angle ABC$ from both sides,we get:
$\angle BAL = \angle ACB$.
Hence,proved.

(N/A) Let the normals at $A$ and $B$ meet at $P$.
Since the mirrors are perpendicular to each other,$BP \parallel OA$ and $AP \parallel OB$.
Thus,$BP \perp PA$,i.e.,$\angle BPA = 90^{\circ}$.
In $\triangle BPA$,by the angle sum property:
$\angle 2 + \angle 3 + \angle BPA = 180^{\circ}$
$\angle 2 + \angle 3 + 90^{\circ} = 180^{\circ}$
$\angle 2 + \angle 3 = 90^{\circ} ......(1)$
By the laws of reflection,the angle of incidence equals the angle of reflection:
$\angle 1 = \angle 2$ and $\angle 3 = \angle 4$
Substituting these into $(1)$:
$\angle 1 + \angle 4 = 90^{\circ} ......(2)$
Now,consider the sum of the angles $\angle CAB$ and $\angle DBA$:
$\angle CAB + \angle DBA = (\angle 1 + \angle 2) + (\angle 3 + \angle 4)$
$= 2\angle 2 + 2\angle 3 = 2(\angle 2 + \angle 3)$
$= 2(90^{\circ}) = 180^{\circ}$
Since the sum of the interior angles on the same side of the transversal $AB$ is $180^{\circ}$,the lines $CA$ and $BD$ must be parallel.
Therefore,$CA \parallel BD$.

(N/A) Consider the triangle $ABC$. By the angle sum property of a triangle,we have:
$\angle A + \angle ABC + \angle ACB = 180^{\circ}$
Dividing the entire equation by $2$,we get:
$\frac{1}{2} \angle A + \frac{1}{2} \angle ABC + \frac{1}{2} \angle ACB = 90^{\circ}$
Since $BO$ and $CO$ are the angle bisectors of $\angle B$ and $\angle C$ respectively,we can write $\angle OBC = \frac{1}{2} \angle ABC$ and $\angle OCB = \frac{1}{2} \angle ACB$. Substituting these into the equation:
$\frac{1}{2} \angle A + \angle OBC + \angle OCB = 90^{\circ} .......(1)$
Now,consider the triangle $OBC$. By the angle sum property of a triangle:
$\angle BOC + \angle OBC + \angle OCB = 180^{\circ} .......(2)$
From equation $(1)$,we have $(\angle OBC + \angle OCB) = 90^{\circ} - \frac{1}{2} \angle A$. Substituting this into equation $(2)$:
$\angle BOC + (90^{\circ} - \frac{1}{2} \angle A) = 180^{\circ}$
Therefore,$\angle BOC = 180^{\circ} - 90^{\circ} + \frac{1}{2} \angle A$
$\angle BOC = 90^{\circ} + \frac{1}{2} \angle A$.

(N/A) Given : $\triangle ABC$,produce $BC$ to $D$ and the bisectors of $\angle ABC$ and $\angle ACD$ meet at point $T$.
To prove : $\angle BTC = \frac{1}{2} \angle BAC$
Proof : In $\triangle ABC$,$\angle ACD$ is an exterior angle.
$\therefore \angle ACD = \angle ABC + \angle CAB$
[Exterior angle of a triangle is equal to the sum of two opposite interior angles]
$\Rightarrow \frac{1}{2} \angle ACD = \frac{1}{2} \angle CAB + \frac{1}{2} \angle ABC$ [Dividing both sides by $2$]
$\Rightarrow \angle TCD = \frac{1}{2} \angle CAB + \frac{1}{2} \angle ABC .....(1)$
[Since $CT$ is a bisector of $\angle ACD$,$\angle TCD = \frac{1}{2} \angle ACD$]
In $\triangle BTC$,$\angle TCD$ is an exterior angle.
$\therefore \angle TCD = \angle BTC + \angle CBT$
[Exterior angle of the triangle is equal to the sum of two opposite interior angles]
$\Rightarrow \angle TCD = \angle BTC + \frac{1}{2} \angle ABC .....(2)$
[Since $BT$ is a bisector of $\angle ABC$,$\angle CBT = \frac{1}{2} \angle ABC$]
From equation $(1)$ and $(2)$,we get
$\frac{1}{2} \angle CAB + \frac{1}{2} \angle ABC = \angle BTC + \frac{1}{2} \angle ABC$
$\Rightarrow \frac{1}{2} \angle CAB = \angle BTC$ or $\angle BTC = \frac{1}{2} \angle BAC$
Hence,proved.

(N/A) Let $AB$ be a given line and $P$ be a point outside the line.
From point $P$,draw a perpendicular $PM$ to the line $AB$,such that $\angle PMB = 90^{\circ}$.
Suppose,if possible,we can draw another perpendicular $PN$ to the line $AB$,such that $\angle PNB = 90^{\circ}$.
In $\triangle PMN$,since $\angle PMB = 90^{\circ}$ and $\angle PNB = 90^{\circ}$,we observe that if $N$ and $M$ are distinct points,then $\triangle PMN$ would have two angles equal to $90^{\circ}$,which implies the sum of angles in the triangle would exceed $180^{\circ}$. This contradicts the angle sum property of a triangle.
Therefore,the points $M$ and $N$ must coincide.
Hence,through a given point,we can draw only one perpendicular to a given line.

(N/A) Given: In $\triangle PQR$,$\angle Q > \angle R$,$PA$ is the bisector of $\angle QPR$ and $PM \perp QR$.
To prove: $\angle APM = \frac{1}{2}(\angle Q - \angle R)$.
Proof:
$1$. Since $PA$ is the bisector of $\angle QPR$,we have $\angle QPA = \angle APR = \frac{1}{2} \angle QPR$.
$2$. In $\triangle PQM$,$\angle Q + \angle PMQ + \angle QPM = 180^{\circ}$. Since $\angle PMQ = 90^{\circ}$,$\angle Q + 90^{\circ} + \angle QPM = 180^{\circ}$,so $\angle Q = 90^{\circ} - \angle QPM$.
$3$. In $\triangle PRM$,$\angle R + \angle PMR + \angle RPM = 180^{\circ}$. Since $\angle PMR = 90^{\circ}$,$\angle R + 90^{\circ} + \angle RPM = 180^{\circ}$,so $\angle R = 90^{\circ} - \angle RPM$.
$4$. Subtracting the two equations: $\angle Q - \angle R = (90^{\circ} - \angle QPM) - (90^{\circ} - \angle RPM) = \angle RPM - \angle QPM$.
$5$. We can write $\angle RPM = \angle APM + \angle APR$ and $\angle QPM = \angle QPA - \angle APM$.
$6$. Substituting these: $\angle Q - \angle R = (\angle APM + \angle APR) - (\angle QPA - \angle APM)$.
$7$. Since $\angle QPA = \angle APR$,this simplifies to $\angle Q - \angle R = 2 \angle APM$.
$8$. Therefore,$\angle APM = \frac{1}{2}(\angle Q - \angle R)$. Hence proved.

(N/A) Since $AB$ is a straight line,$\angle APC + \angle BPC = 180^{\circ}$ (Linear pair of angles).
Given that $\angle APC : \angle BPC = 7 : 8$.
Let $\angle APC = 7x$ and $\angle BPC = 8x$.
Then,$7x + 8x = 180^{\circ}$.
$15x = 180^{\circ}$.
$x = \frac{180^{\circ}}{15} = 12^{\circ}$.
Therefore,$\angle APC = 7 \times 12^{\circ} = 84^{\circ}$ and $\angle BPC = 8 \times 12^{\circ} = 96^{\circ}$.
Since $AB$ and $CD$ are straight lines intersecting at $P$,the vertically opposite angles are equal.
Thus,$\angle DPB = \angle APC = 84^{\circ}$ and $\angle APD = \angle BPC = 96^{\circ}$.
So,the four angles are $84^{\circ}, 96^{\circ}, 84^{\circ},$ and $96^{\circ}$.

(A) Since ray $YW$ stands on line $XYZ$,$\angle WYZ + \angle WYX = 180^{\circ}$ (Linear pair of angles).
Given that $\angle WYZ : \angle WYX = 1 : 2$,let $\angle WYZ = x$ and $\angle WYX = 2x$.
Therefore,$x + 2x = 180^{\circ} \implies 3x = 180^{\circ} \implies x = 60^{\circ}$.
So,$\angle WYZ = 60^{\circ}$ and $\angle WYX = 120^{\circ}$.
Ray $YQ$ is the angle bisector of $\angle WYZ$,so $\angle WYQ = \frac{1}{2} \times \angle WYZ = \frac{1}{2} \times 60^{\circ} = 30^{\circ}$.
Ray $YP$ is the angle bisector of $\angle WYX$,so $\angle WYP = \frac{1}{2} \times \angle WYX = \frac{1}{2} \times 120^{\circ} = 60^{\circ}$.
Finally,$\angle PYQ = \angle WYQ + \angle WYP = 30^{\circ} + 60^{\circ} = 90^{\circ}$.

(N/A) Given that $AB \perp CD$,therefore $\angle ABC = 90^{\circ}$.
From the figure,$\angle ABC = x + y + z = 90^{\circ}$.
The ratio of the angles is given as $x : y : z = 4 : 5 : 6$.
Let the angles be $4k, 5k,$ and $6k$ respectively.
Sum of the ratios $= 4 + 5 + 6 = 15$.
Therefore,$15k = 90^{\circ}$,which gives $k = \frac{90^{\circ}}{15} = 6^{\circ}$.
Now,calculating the values:
$x = 4k = 4 \times 6^{\circ} = 24^{\circ}$
$y = 5k = 5 \times 6^{\circ} = 30^{\circ}$
$z = 6k = 6 \times 6^{\circ} = 36^{\circ}$
Thus,the values are $x = 24^{\circ}, y = 30^{\circ},$ and $z = 36^{\circ}$.

(N/A) Given that lines $AB$ and $CD$ intersect at $O.$
Since $AB$ is a straight line,the sum of angles on one side of the line is $180^{\circ}.$
Thus,$\angle AOC + \angle COE + \angle BOE = 180^{\circ}.$
We are given $\angle AOC + \angle BOE = 100^{\circ}.$
Substituting this into the equation: $100^{\circ} + \angle COE = 180^{\circ},$
which gives $\angle COE = 180^{\circ} - 100^{\circ} = 80^{\circ}.$
Since $AB$ and $CD$ are straight lines,vertically opposite angles are equal.
Therefore,$\angle AOC = \angle BOD = 40^{\circ}.$
Using the given equation $\angle AOC + \angle BOE = 100^{\circ},$
$40^{\circ} + \angle BOE = 100^{\circ},$
$\angle BOE = 100^{\circ} - 40^{\circ} = 60^{\circ}.$
Finally,the reflex $\angle COE = 360^{\circ} - \angle COE = 360^{\circ} - 80^{\circ} = 280^{\circ}.$

(N/A) Given that ray $QS \perp PR$,therefore $\angle PQS = \angle SQR = 90^{\circ}$.
Since $\angle PQT + \angle TQS = \angle PQS$,we have $b + a = 90^{\circ}$.
Given $a: b = 7: 11$,let $a = 7x$ and $b = 11x$.
Substituting these into the equation: $7x + 11x = 90^{\circ} \implies 18x = 90^{\circ} \implies x = 5^{\circ}$.
Thus,$a = 7 \times 5^{\circ} = 35^{\circ}$ and $b = 11 \times 5^{\circ} = 55^{\circ}$.
Since $PR$ is a straight line,$\angle PQT + \angle TQR = 180^{\circ}$ (linear pair).
$b + \angle TQR = 180^{\circ} \implies 55^{\circ} + \angle TQR = 180^{\circ} \implies \angle TQR = 125^{\circ}$.
Also,$\angle TQR$ and $\angle PQU$ are vertically opposite angles,so $\angle PQU = 125^{\circ}$.
Since $\angle PQU = \angle PQT + \angle TQU$,and $\angle PQT = b = 55^{\circ}$,then $\angle TQU = 125^{\circ} - 55^{\circ} = 70^{\circ}$.
From the figure,$\angle RQU = c$ and $\angle PQT$ and $\angle RQU$ are vertically opposite angles? No,$PR$ and $TU$ are lines intersecting at $Q$.
Thus,$\angle PQT = \angle RQU = b = 55^{\circ}$ and $\angle PQU = \angle TQR = 125^{\circ}$.
Therefore,$c = 55^{\circ}$ (vertically opposite to $b$).
Reflex $\angle TQR = 360^{\circ} - \angle TQR = 360^{\circ} - 125^{\circ} = 235^{\circ}$.

(A) Draw a line $PQ \parallel AB$,passing through point $X$ as shown in the figure.
Now,$PQ \parallel AB$ and $AB \parallel CD$.
Therefore,$CD \parallel PQ$ (since $AB \parallel CD$ and $AB \parallel PQ$).
Consider the transversal $MX$ intersecting parallel lines $AB$ and $PQ$. The interior angles on the same side of the transversal are supplementary.
$\therefore \angle BMX + \angle MXQ = 180^{\circ}$
Given $\angle BMX = 125^{\circ}$,we have:
$125^{\circ} + \angle MXQ = 180^{\circ}$
$\therefore \angle MXQ = 180^{\circ} - 125^{\circ} = 55^{\circ}$ $(1)$
Now,consider the transversal $NX$ intersecting parallel lines $CD$ and $PQ$. The alternate interior angles are equal.
$\therefore \angle NXQ = \angle XNC = 55^{\circ}$ $(2)$
Finally,$\angle MXN = \angle MXQ + \angle NXQ$.
Substituting the values from $(1)$ and $(2)$:
$\angle MXN = 55^{\circ} + 55^{\circ} = 110^{\circ}$.

$(96^{\circ})$ Given: $PQ \parallel RS$ and $RS \parallel TU$.
Since lines parallel to the same line are parallel to each other,we have $PQ \parallel TU$.
Therefore,$x = z$ (Alternate interior angles).
Now,for $PQ \parallel RS$,the interior angles on the same side of the transversal are supplementary.
Therefore,$x + y = 180^{\circ}$.
Substituting $x = z$ in the equation,we get $z + y = 180^{\circ}$.
Given the ratio $y: z = 7: 8$,let $y = 7k$ and $z = 8k$.
Substituting these into the equation: $7k + 8k = 180^{\circ} \implies 15k = 180^{\circ} \implies k = 12^{\circ}$.
Thus,$y = 7 \times 12^{\circ} = 84^{\circ}$ and $z = 8 \times 12^{\circ} = 96^{\circ}$.
Since $x = z$,we have $x = 96^{\circ}$.

(D) Given that $PQ || RS$ and $RS || MN$. Therefore,$PQ || MN$.
From the figure,the angle $x$ and $z$ are alternate interior angles if we consider the transversal line,but looking at the positions,$x$ and $z$ are interior angles on the same side of the transversal for lines $PQ$ and $MN$,so $x + z = 180^o$.
Also,$y$ and $z$ are consecutive interior angles between lines $RS$ and $MN$,so $y + z = 180^o$.
This implies $y = x$ (as both are supplementary to $z$).
Given $y: z = 7: 11$,let $y = 7k$ and $z = 11k$.
Since $y + z = 180^o$,we have $7k + 11k = 180^o$,which gives $18k = 180^o$,so $k = 10^o$.
Thus,$y = 7 \times 10^o = 70^o$ and $z = 11 \times 10^o = 110^o$.
Since $x + z = 180^o$ (consecutive interior angles for $PQ || MN$),we have $x + 110^o = 180^o$.
Therefore,$x = 180^o - 110^o = 70^o$.
Wait,checking the figure again: $x$ and $y$ are consecutive interior angles between $PQ$ and $RS$,so $x + y = 180^o$. Since $y = 70^o$,$x = 180^o - 70^o = 110^o$.

(N/A) $1$. Given: $PQ \parallel RS$,$YZ \perp RS$ (i.e.,$\angle YZS = 90^{\circ}$),and $\angle SZX = 143^{\circ}$.
$2$. Since $RS$ is a straight line,$\angle RZX + \angle SZX = 180^{\circ}$ (Linear pair).
$\angle RZX + 143^{\circ} = 180^{\circ} \implies \angle RZX = 37^{\circ}$.
$3$. Since $PQ \parallel RS$,the alternate interior angles are equal. Thus,$\angle PXZ = \angle RZX = 37^{\circ}$.
$4$. Since $YZ \perp RS$,$\angle YZS = 90^{\circ}$. Also,$\angle XZY + \angle YZS = \angle XZS$. Since $\angle XZS$ and $\angle RZX$ are vertically opposite angles,$\angle XZS = \angle RZX = 37^{\circ}$. However,looking at the geometry,$\angle XZY = \angle YZS - \angle XZS = 90^{\circ} - 37^{\circ} = 53^{\circ}$.
$5$. In $\triangle XYZ$,the sum of angles is $180^{\circ}$. $\angle YXZ + \angle XZY + \angle XYZ = 180^{\circ}$. Since $PQ \parallel RS$ and $YZ \perp RS$,$\angle XYZ = 90^{\circ}$.
$\angle YXZ + 53^{\circ} + 90^{\circ} = 180^{\circ} \implies \angle YXZ = 180^{\circ} - 143^{\circ} = 37^{\circ}$.

(B) To find $\angle BPC$,draw a line $EF$ passing through point $P$ such that $EF \parallel AB$. Since $AB \parallel CD$ and $EF \parallel AB$,it follows that $EF \parallel CD$.
$1$. Consider the parallel lines $AB$ and $EF$. Since $BP$ is a transversal,the interior angles on the same side sum to $180^{\circ}$.
$\angle PBA + \angle BPE = 180^{\circ}$
$140^{\circ} + \angle BPE = 180^{\circ}$
$\angle BPE = 180^{\circ} - 140^{\circ} = 40^{\circ}$.
$2$. Consider the parallel lines $CD$ and $EF$. Since $CP$ is a transversal,the interior angles on the same side sum to $180^{\circ}$.
$\angle PCD + \angle CPE = 180^{\circ}$
$130^{\circ} + \angle CPE = 180^{\circ}$
$\angle CPE = 180^{\circ} - 130^{\circ} = 50^{\circ}$.
$3$. Now,$\angle BPC = \angle BPE + \angle CPE = 40^{\circ} + 50^{\circ} = 90^{\circ}$.

(C) Given $AB \parallel CD$.
Since $AB \parallel CD$ and $XY$ is a transversal,the alternate interior angles are equal. Therefore,$\angle C Y X = \angle AXY = 80^{\circ}$.
Since $CYD$ is a straight line,$\angle C Y X + \angle X Y Z = 180^{\circ}$ (linear pair).
$80^{\circ} + x = 180^{\circ} \implies x = 100^{\circ}$.
Wait,looking at the figure,$\angle X Y Z$ is an interior angle. Since $AB \parallel CD$,$\angle AXY + \angle X Y C = 180^{\circ}$ (consecutive interior angles).
$80^{\circ} + \angle X Y C = 180^{\circ} \implies \angle X Y C = 100^{\circ}$.
Since $CYD$ is a straight line,$\angle X Y C + \angle X Y Z = 180^{\circ} \implies 100^{\circ} + x = 180^{\circ} \implies x = 80^{\circ}$.
Now,for $\angle XZD = 140^{\circ}$,since $CYD$ is a straight line,$\angle X Z Y + \angle X Z D = 180^{\circ} \implies \angle X Z Y + 140^{\circ} = 180^{\circ} \implies \angle X Z Y = 40^{\circ}$.
In $\triangle XYZ$,the sum of angles is $180^{\circ}$.
$\angle X Y Z + \angle X Z Y + \angle Y X Z = 180^{\circ} \implies 80^{\circ} + 40^{\circ} + y = 180^{\circ} \implies 120^{\circ} + y = 180^{\circ} \implies y = 60^{\circ}$.
Thus,$x = 80^{\circ}$ and $y = 60^{\circ}$.

(N/A) In $\Delta ABC$,the sum of angles is $180^{\circ}$,so $\angle A + \angle ABC + \angle ACB = 180^{\circ}$.
$\therefore \angle ABC + \angle ACB = 180^{\circ} - \angle A \quad \dots(1)$
The bisectors of $\angle B$ and $\angle C$ intersect at $I$.
$\therefore \angle IBC = \frac{1}{2} \angle ABC$ and $\angle ICB = \frac{1}{2} \angle ACB$.
In $\Delta IBC$,the sum of angles is $180^{\circ}$,so $\angle BIC + \angle IBC + \angle ICB = 180^{\circ}$.
$\therefore \angle BIC = 180^{\circ} - (\angle IBC + \angle ICB)$.
Substituting the values of $\angle IBC$ and $\angle ICB$:
$\therefore \angle BIC = 180^{\circ} - \left[ \frac{1}{2} \angle ABC + \frac{1}{2} \angle ACB \right]$.
$\therefore \angle BIC = 180^{\circ} - \frac{1}{2} (\angle ABC + \angle ACB)$.
Using equation $(1)$:
$\therefore \angle BIC = 180^{\circ} - \frac{1}{2} (180^{\circ} - \angle A)$.
$\therefore \angle BIC = 180^{\circ} - 90^{\circ} + \frac{1}{2} \angle A$.
$\therefore \angle BIC = 90^{\circ} + \frac{1}{2} \angle A$.

(N/A) In $\Delta ABC$,$\angle ACD$ is an exterior angle. By the exterior angle theorem,$\angle ACD = \angle BAC + \angle ABC$.
In $\Delta EBC$,$\angle ECD$ is an exterior angle. Therefore,$\angle ECD = \angle EBC + \angle BEC$.
Since $BE$ is the bisector of $\angle ABC$,$\angle EBC = \frac{1}{2} \angle ABC$.
Since $CE$ is the bisector of $\angle ACD$,$\angle ECD = \frac{1}{2} \angle ACD = \frac{1}{2} (\angle BAC + \angle ABC)$.
Substituting these into the equation for $\Delta EBC$: $\frac{1}{2} (\angle BAC + \angle ABC) = \frac{1}{2} \angle ABC + \angle BEC$.
Simplifying,$\frac{1}{2} \angle BAC + \frac{1}{2} \angle ABC = \frac{1}{2} \angle ABC + \angle BEC$.
Thus,$\angle BEC = \frac{1}{2} \angle BAC$.