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#### In Fig. 6.17, $\angle Q > \angle R$, PA is the bisector of $\angle QPR$ and $PM \perp QR$. Prove that  $\angle APM = \frac{1}{2} (\angle Q -\angle R)$.

Solution:

Given : In $\triangle PQR$, $\angle Q > \angle R$

PA is the bisector of $\angle QPR$  and $PM \perp QR$.

To prove : $\angle APM =\left ( \frac{1}{2} \right ) (\angle Q -\angle R)$
Proof: Since PA is the bisector of $\angle P$, we have

$\angle APQ = \left ( \frac{1}{2} \right )\angle P$                        … (i)
In right angled $\triangle PMQ$ we have

$\angle Q + \angle MPQ + 90^{\circ} = 180^{\circ}$ (Angle sum property)

$\Rightarrow \angle MPQ = 90^{\circ} - \angle Q$              … (ii)
Now
$\angle APM = \angle APQ - \angle MPQ$
$\angle APM =\frac{1}{2} \angle P - (90 - \angle Q)$         using (i) & (ii)
$\angle APM = \frac{1}{2} \angle P - 90^{\circ} + \angle Q$

$\angle APM = \frac{1}{2}\angle P -\frac{1}{2} (\angle P + \angle R + \angle Q) + \angle Q$  Since $90^{\circ} = \frac{1}{2}(\angle P + \angle R + \angle Q)$
$\angle APM = \frac{1}{2}\angle P -\frac{1}{2} \angle P - \frac{1}{2} \angle R-\frac{1}{2} \angle Q + \angle Q$
$\angle APM =\frac{1}{2} (\angle Q -\angle R)$

Hence proved

#### Prove that a triangle must have at least two acute angles.

Solution.

It is given that a triangle must have at least two acute angles.
An acute angle is less than 90 degrees
Let us assume that a triangle does not have two acute angles.

So, it has two angles that are either right angles (=90 degrees) or obtuse angles (greater than 90 degrees)
So let two right angles are present,
So using angle sum property of a triangle, third angle must be zero which is not possible.

Also let one angle be right and one be obtuse. We can take the smallest obtuse angle, i.e.,
So using angle sum property of a triangle, third angle must be negative which is not possible.

Again, if both the angles are obtuse the third angle must be negative which is not possible.
So it is necessary for a triangle to have at least two acute angles.

Hence proved

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#### Prove that two lines that are respectively perpendicular to two intersecting lines intersect each other. [Hint: Use proof by contradiction].

Given: Let line x and y are two intersecting lines. Let n and p be another two lines which are perpendicular to x and y

To prove: n and p intersect at a point
Proof: Let lines n and p are not intersecting then $n \parallel p$ … (1)
Since n and p are parallel and n is perpendicular to x and p is perpendicular to y respectively

So, $x \parallel y$
But, it is a contradiction as it is given that x and y are two intersecting lines

Thus our assumption is wrong.
n and p intersect at a point

Hence proved

#### Prove that through a given point, we can draw only one perpendicular to a given line. [Hint: Use proof by contradiction].

Solution:

Given: Consider a line R and a point P

Construction:
Draw two lines (m and n) passing through P which are perpendicular to line R.
To prove: Only one perpendicular line can be drawn through a point P
Proof: In $\triangle APB$
$\angle A + \angle P + \angle B = 180^{\circ}$   {angle sum property}

$90^{\circ} + \angle P + 90^{\circ} = 180^{\circ}$
$\angle P = 180 - 180^{\circ}$
$%u200B%u200B%u200B%u200B%u200B%u200B%u200B\angle P = 0^{\circ}$
So lines n and m will coincide
Therefore we can draw only one perpendicular to a given line.

Hence proved

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#### A transversal intersects two parallel lines. Prove that the bisectors of any pair of corresponding angles so formed are parallel.

Given: A transversal EF cuts two parallel line AB and CD at point G & H. GL and HM are bisectors of angles

To prove: $GL \parallel HM$
Proof: $\angle EGB = \angle GHD$ (Corresponding angles)
$\frac{1}{2} \angle EGB =\frac{1}{2} \angle GHD$

$\angle EGL = \angle GHM$
These are the corresponding angle formed by the line GL and HM, where EF is the transversal.

$\therefore GL \parallel HM$

Hence proved

#### Bisectors of interior $\angle B$ and exterior $\angle ACD$ of a $\triangle ABC$ intersect at the point T. Prove that $\angle BTC = \frac{1}{2} \angle BAC$.

According to the question,
Bisectors of interior $\angle B$ and exterior $\angle ACD$ of a $\triangle ABC$ intersect at the point T
$\angle TBC =\frac{1}{2} \angle ABC$                          … (1)
And $\angle TCD =\frac{1}{2} \angle ACD$                  … (2)

Now from $\triangle ABC$ we have
$\angle BAC + \angle ABC = \angle ACD$ … (3) (exterior angle is equal to the sum of interior opposite angles)
And from $\triangle TBC$ we have
$\angle BTC + \angle TBC = \angle TCD$ (exterior angle is equal to the sum of interior opposite angles)
Or $\angle BTC + \frac{1}{2} \angle ABC =\frac{1}{2} \angle ACD$ using (1 and 2)
Or $\angle BTC = \frac{1}{2}(\angle ACD - \angle ABC)$
Using (3), $\angle ACD - \angle ABC = \angle BAC$
So,

$\angle BTC = \frac{1}{2} \angle BAC$

Hence Proved

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#### If two lines intersect, prove that the vertically opposite angles are equal.

Solution.

It is given that if two lines intersect, the vertically opposite angles are equal.
Proof:

Now let AB and CD be two lines intersecting at point O.

From the figure, we have two pairs of vertically opposite angles namely:

(i) $\angle AOC$ and $\angle BOD$

(ii) $\angle AOD$ and $\angle BOC$

Now we have to prove that $\angle AOC = \angle BOD$

And $\angle AOD = \angle BOC$

$\Rightarrow$ Now ray OA stands on line CD

$\therefore \angle AOC + \angle AOD = 180^{\circ}$ … (i) (linear pair angles)

Similarly, can we write

$\angle AOD + \angle BOD = 180^{\circ}$ … (ii) (linear pair angles)

From equation (i) and (ii) comparing

$\angle AOC + \angle AOD = \angle AOD + \angle BOD$

$\Rightarrow \angle AOC = \angle BOD$

Similarly, we can prove that $\angle AOD = \angle BOC$

Hence Proved.

#### Two lines are respectively perpendicular to two parallel lines. Show that they are parallel to each other.

Given: Two lines m and n are parallel and another two lines p and q are respectively perpendicular to m and n, i.e., $p \perp m, p \perp n, q \perp m, q \perp n$.
To prove
$p \parallel q$

Proof:

Since m||n and p is perpendicular to m and n.
$\angle 5 = \angle 6 = \angle 7 = \angle 8 = \angle 11 = \angle 12 = 90^{\circ}$
Similarly, q is perpendicular to m and n.
$�\\angle 1 = \angle 2 = \angle 3 = \angle 4 = \angle 9 = \angle 10 = 90^{\circ}$
Now for lines p and q, m is the transversal
$\angle 1 = \angle 2 = \angle 3 = \angle 4 = \angle 5 = \angle 6 = \angle 7 = \angle 8 = 90^{\circ}$
So we can see that all the conditions are fulfilled for the lines to be parallel, i.e., Corresponding angles are equal, sum of cointerior angles is 180o, alternate angles are equal.

Hence, $p \parallel q$

Hence proved

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#### A triangle ABC is right angled at A. L is a point on BC such that $AL\perp BC$. Prove that $\angle BAL = \angle ACB$.

Solution:$AL\perp BC$ $\angle BAL = \angle ACB$

Proof: In $\triangle ABC$ and $\triangle LAB$,
$\angle BAC = \angle ALB$ (each $90^{\circ}$) (i)
And $\angle ABC = \angle ABL$ (Common angle) (ii)
In $\triangle ABC$,
$\angle BAC + \angle ABC + \angle ACB = 180^{\circ}$ (angle sum property)
$\angle ABC + \angle ACB = 180^{\circ} -90^{\circ}$ (from i)
$\angle ACB = 90^{\circ} - \angle ABC$
In $\triangle ABL$,
$\angle BAL + \angle ALB + \angle ABL = 180^{\circ}$(angle sum property)
$\angle BAL + \angle ABL = 180^{\circ} - 90^{\circ}$ (from i)
$\angle BAL = 90^{\circ} - \angle ABL$
$\angle BAL = 90^{\circ} - \angle ABC$ (from ii)
Hence, $\angle ACB = 90^{\circ} -\angle ABC = \angle BAL$

Hence proved

#### The angles of a triangle are in the ratio 2 : 3 : 4. Find the value of each angle. What type of triangle is it?

Angles of the triangle are in the ratio- 2 : 3 : 4
Let the angles are 2x, 3x, 4x then:
$2x + 3x + 4x = 180^{\circ}$
(angle sum property)
$9x = 180^{\circ}$
$x = 20^{\circ}$
Then the angles of the triangle are:
$2x = 40^{\circ}$
$3x = 60^{\circ}$
$4x = 80^{\circ}$
This triangle is a scalene triangle as all the angles are of different measure.