Notesale: Turn your study into money

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Triangles
Medians of Triangles
Let us consider the following triangle ABC
...
A line segment
the mid-point of
Then, we say that

is the side opposite to

is drawn joining the point A and the point D, where D is


...


A median can be defined as follows
...
”
Now we know what a median is, can we tell how many medians can be drawn inside a
triangle?
In a triangle, there are three vertices
...


Here,

, and

are the three medians of ΔABC
...

From the figure, it can be observed that the medians
other at a common point G
...
"
Thus, medians of a triangle are concurrent
...

In the above given figure, G is the point of concurrence
...
Draw a △ABC
...
With B and C as centres and radius more than half of BC, draw two arcs intersecting
at points X and Y
...


3
...
Join MN thus meeting the line AC at point Q
...
Similarly, draw the perpendicular bisector of line AB meeting AB at point R
...
Join AP, BQ and CR
...


Point O is the centroid of △△ABC and AP, BQ and CR are the medians of sides BC, AC
and AB respectively
...

Example 1:
In the triangle PQR, PS is a median and the length of
of


...
5 cm
...
Therefore, S is the mid-point of QR
...
5 cm
= 13 cm

Altitudes of Triangles
The students of class VII were being taken to a tour to Corbett National Park
...
The entrance in the tent was of a triangular shape as shown in
the following figure
...

Similarly, in any triangle, we can draw a perpendicular which represents its height
...


Look at the triangle PQR below
...


is the opposite side of the vertex P
...
Line segment

is a

is called the height or altitude of

An altitude can be defined as follows
...
”
Note: A triangle can have three altitudes
...


"The point of intersection of the altitudes is called the orthocentre of the triangle
...
Using set-square

1
...


2
...
Where the perpendicular meets the
side PQ, name it as point X
...


3
...

Thus, the altitudes obtained are XR, QZ and PY
...
Using Compasses
1
...

2
...

3
...
Draw a line
joining XY cutting the line QR at point M
...


4
...
With C
and D as centres, draw two intersecting arcs
...
Join
QN
...
Similarly, draw altitude from point R on PQ cutting the line PQ at point L
...
Join PM, RL and QN and name the meeting point of these three altitudes as O
...

Remember
The altitudes of a triangle may not always lie inside it
...
In this case, we have to extend the opposite side of the
vertex from which the altitude is drawn
...

is the altitude of ΔABC drawn from the vertex A to extended side

...
And,
is the
altitude drawn from B to

...

It can be seen that altitudes in each triangle intersect each other at a common point
...

In ΔPQR and ΔABC, G is the point of concurrence
...

Example:
In triangle ABC,

is perpendicular to

and the altitude drawn from A to

such that

=


...


Therefore,

is a median of ΔABC drawn from the vertex A to

Thus, the altitude and the median drawn from A to


...


Angle Sum Property of Triangles
If we join any three non-collinear points in a plane, then we get a triangle
...


The sum of the three interior angles of a triangle is 180° and this property of a triangle
is known as the angle sum property
...
e
...

The angle sum property was identified by the Pythagorean school of Greek
mathematicians (or the Pythagoreans) and proved by Euclid
...

Proving the Angle Sum Property of Triangles
Know Your Scientist
Pythagoras

Pythagoras (570 BC−495 BC) was a great Greek mathematician
and philosopher, often described as the first pure mathematician
...
He also made
influential contributions to philosophy and religious teaching
...
This
society followed a code of secrecy, which is the reason why a
sense of mystery surrounds the figure of Pythagoras
...
He is referred to as ‘the father of geometry’
...
Euclid wrote a series of books which are
collectively known as the Elements
...

The Elements served as the main textbook for teaching
mathematics (especially geometry) from the time of its publication
up until the early 20th century
...


Did You Know?
In 1942, a Dutch mathematics teacher Albert E
...
He named it the Pythagoras tree because of the presence of
right-angled triangles in the figure
...
This fact can be proved
as is shown
...

According to the angle sum property, we have:
∠A + ∠B + ∠C = 180°

⇒ 90° + 90° + ∠C = 180°
⇒ ∠C = 180° − 180°
⇒ ∠C = 0°
However, the above is not possible
...

Similarly, we can prove that a triangle cannot have two obtuse angles
...

• The smallest interior angle is opposite the smallest side
...

Facts about the Angle Sum Property
By the angle sum property, we can deduce the fact that there can be no triangle with
all angles less than or greater than 60°
...

Consider a ΔABC with all angles equal to 59°
...

By adding the given angles, we obtain:
59° + 59° + 59° = 177° ≠ 180°
Since ΔABC does not satisfy the angle sum property, it cannot exist
...

According to the angle sum property, we should have ∠A + ∠B + ∠C = 180°
...


Thus, we have proved that a triangle cannot have all angles less than or greater
than 60°
...


Solution:
We know that the sum of the three angles of a triangle is 180°
...

Let the angles be x, 3x and 5x
...

Medium
Example 1:
In the given ΔXYZ, YO, ZO and PO are the respective bisectors of ∠XYZ, ∠XZY
and ∠YOZ
...


Solution:
As per the relation between the vertex angle and the angles made by the bisectors of
the remaining two angles, we have:

In ΔOYP, we have:

∠YOP =

⇒ ∠YOP =

∠YOZ (&because PO is the bisector of ∠YOZ)

× 122°

⇒ ∠YOP = 61°
Again in ΔOYP, we have:
∠YOP + ∠OPY + ∠PYO = 180° (By the angle sum property)
⇒ 61° + 91° + ∠PYO = 180°
⇒ ∠PYO = 180° − 152°
⇒ ∠PYO = 28°
We know that YO is the bisector of ∠XYZ
...


Solution:
Using the angle sum property in ΔRST, we obtain:
∠RST + ∠RTS + ∠SRT = 180°
⇒ 61° + 73° + ∠SRT = 180°
⇒ 134° + ∠SRT = 180°
⇒ ∠SRT = 180° − 134°
⇒ ∠SRT = 46°
In ΔRST and ΔRPQ, we have:
∠PRQ = ∠SRT = 46° (Vertically opposite angles)
Using the angle sum property in ΔRPQ, we obtain:
∠RPQ + ∠RQP + ∠PRQ = 180°
⇒ x° + x° + 46° = 180°
⇒ 2x° + 46° = 180°
⇒ 2x° = 180° − 46°
⇒ 2x° = 134°
⇒ x = 67

Hard
Example 1:
In the figure, l and m are two plane mirrors placed perpendicular to each other
...


Solution:
It is given that mirrors l and m are perpendicular to each other
...
Mark the
angles made by these perpendiculars with the incident and reflected rays as is shown
...

∴ ∠1 = ∠2
For mirror m, ∠3 is the angle of incidence and ∠4 is the angle of reflection
...
Therefore, the
incident ray CA is parallel to the reflected ray BD
...
Prove that ∠GLH = 90°
...
e
...

∴ GM||HL
Similarly, we can prove that GL||HM
...

We know that AB||CD and EF is the transversal
...


In the given figure, an open angle formed by the edge of the triangular structure with the
horizontal plane is marked
...
Such angles are known
as exterior angles
...

Exterior Angles of Triangles
Look at the triangle shown
...
This extended side
forms an angle with side AB, i
...
, ∠ABD
...

Hence, ∠ABD is an exterior angle of ΔABC
...

It can be seen that exterior ∠ABD forms linear pair with interior ∠ABC of ΔABC
...

Such angles are known as the remote interior angles of an exterior angle
...

Exterior Angle Theorem or Remote Interior Angles Theorem and Its Proof
Corollary Related to Exterior Angle Theorem
There is a corollary related to exterior angle theorem which states that:
An exterior angle is always greater than each of its remote interior angles
...


Here, ∠ABD is an exterior angle of the triangle and its interior opposite or remote
interior angles are ∠ACB and ∠CAB
...

Thus, 0° <∠ABC < 180°, 0° < ∠ACB < 180° and 0° < ∠CAB < 180°
Now, ∠ABC < 180°
∴ 180° – ∠ABC > 0°
⇒ ∠ABD > 0°

In triangle ΔABC, we have
∠ABD = ∠ACB +∠CAB

(By exterior angle theorem)

And, ∠ABD > 0, ∠ACB > 0° and ∠CAB > 0°

(Property of triangle)

Therefore, ∠ABD > ∠ACB and ∠ABD >∠CAB
Thus, an exterior angle is always greater than each of its remote interior angles
...
These exterior angles are always of equal measure
...


The figure clearly shows that two exterior angles can be drawn at vertex C—one by
producing BC up to point D and the other by producing AC up to point E
...

According to the exterior angle theorem, the measure of an exterior angle of a triangle is
equal to the sum of the measures of the two opposite interior angles of the triangle
...
The two
angles thus drawn have an equal measure and are equal to the sum of the two opposite
interior angles
...


Solution:
According to the exterior angle property of triangles, the measure of an exterior angle of
a triangle is equal to the sum of the measures of the two opposite interior angles of the
triangle
...


Solution:
∠ABC and ∠ACB are interior angles opposite to the exterior angle at vertex A, i
...
,
∠BAD
...
Find the sum of the three exterior angles so formed
...

Example 2:
Show that AC is the bisector of ∠BAD in the given figure
...
We have found ∠BAC = ∠CAD = 50°
...

Hard
Example 1:
If AD||BE in the given figure, then find the values of a, b, x and y
...

∠ADF = ∠DCB (Corresponding angles)
⇒ 150° = 72° + y
⇒ y = 78° … (1)
Now, y + 72° + a = 180° (As they form line BCE)
⇒ 78° + 72° + a = 180° (Using equation 1)
⇒ a = 180° − 150°
⇒ a = 30° … (2)
Consider ΔCDE
...

∠ABH = x + y (Exterior angle property)
⇒ 126° = x + 78°
⇒ x = 48°
Hence, a = 30°, b = 90°, x = 48° and y = 78°
...
Find the values
of a, b, c and d
...

∴ 126° + c = 180°
⇒ c = 180° − 126°
⇒ c = 54°
On using the exterior angle property in ΔAED, we get:
126° = 72° + d
⇒ d = 126° − 72°
⇒ d = 54°

Since EBCD is a trapezium, BC is parallel to ED
...

So, we have:
a = c = 54° (Pair of corresponding angles)
b = d = 54° (Pair of corresponding angles)
Thus, a = b = c = d = 54°
...

Now, let us study about a special type of triangle known as isosceles triangle
...
”

In ΔABC, two sides AB and AC are of equal length
...

“In an isosceles triangle, angles opposite to equal sides are also equal
...

∴ ∠B = ∠C
Therefore, we can also define isosceles triangle in terms of angles
...
”

In the given figure,
∠Q = ∠R
∴ ΔPQR is an isosceles triangle
...

Example 1:
Find the value of x for the following figures
...


The angles opposite to equal sides are equal
...

The angles opposite to equal sides are equal
...


Solution:
It is given that ΔABC is an isosceles triangle
...

∴ ∠ABC = ∠ACB (1)

Now, ∠ABC = 180° − ∠ABX (By linear pair axiom)
= 180° − 130°
= 50°
From equation (1), we obtain
∠ACB = 50°
Using angle sum property in ΔABC, we obtain
∠ABC + ∠ACB + ∠BAC = 180°
50° + 50° + ∠BAC = 180°
∠BAC = 180° − 100°
∠BAC = 80°
Thus, the measure of ∠BAC is 80°
...


What do you observe about the sides of the hanger? The base is clearly the longest
side, while the other two sides are equal
...
So, the hanger is a real-life example of an
isosceles triangle
...
Yet
another example is a diagonally cut slice of bread
...
In
this lesson, we will discuss about the same, i
...
, angles opposite equal sides of an
isosceles triangle
...


Angles Opposite Equal Sides of an Isosceles Triangle Are Equal
Know Your Scientist
Thales (624 BC−546 BC) is believed to be the first
philosopher in the Greek tradition
...

Thales is credited for giving the first known mathematical
proof, i
...
, ‘a circle is bisected by its diameter’
...
For this reason, this theorem is
called Thales Theorem
...
He also propounded a few general notions such as ‘all
straight angles are equal’, ‘equals subtracted from equals are equal’ and ‘equals added
to equals are equal’
...

Did You Know?
The word ‘isosceles’ is a combination of the Greek words ‘isos’ meaning ‘equal’ and
‘skelos’ meaning ‘leg’
...

Know More
•

The altitude to the base of an isosceles triangle bisects the vertex angle
...


•

When the equal sides are at right angle, the triangle is called a ‘right isosceles triangle’
...

Orthocentre: It is the point where the altitudes of the three sides of a triangle intersect
...

Centroid: It is the point where the medians of the three sides of a triangle intersect
...

Incentre: It is the point where the interior angle bisectors of a triangle intersect
...

Circumcentre: It is the point where the perpendicular bisectors of the three sides of a
triangle intersect
...

Solved Examples
Easy
Example 1:
In the given ΔABC, AB = AC
...

By the property of isosceles triangles, we obtain:

∠ABC = ∠ACB
⇒ ∠ACB = 50°
By the angle sum property, we have:
∠ABC + ∠ACB + ∠BAC = 180°
⇒ 50° + 50° + ∠BAC = 180°
⇒ ∠BAC = 180° − 100°
⇒ ∠BAC = 80°
Example 2:
Ram lives in a triangular tree house
...
He wants to cover the slant sides with metal sheets
...

Solution:
It is given that the base angles are equal; so, the triangular structure is isosceles
...

Now, we know that:
Base = 4 m
Boundary of the triangular structure = Perimeter of the isosceles triangle = 10 m
⇒ 10 m = 4 m + 2 × (Measure of each slant side)
⇒ 2 × (Measure of each slant side) = 10 m − 4 m
⇒ 2 × (Measure of each slant side) = 6 m
Hence, Ram needs 6 m of metal sheet to cover the slant sides of the triangular
structure
...


Solution:
1
...
5 cm
Since the angles opposite equal sides of a triangle are equal, we obtain:
∠ABC = ∠ACB
Let ∠ABC = ∠ACB = x
By the angle sum property of triangles, we have:
∠ABC + ∠ACB + ∠BAC = 180°
⇒ x + x + 80° = 180°
⇒ 2x = 100°
⇒ x = 50°
Thus, ∠ABC = ∠ACB = 50°
2
...
Find the measures of ∠BAC, ∠ABC and
∠ACB
...

∴ ∠ACB + ∠ACD = 180°
⇒ x + 2x = 180°
⇒ 3x = 180°
⇒ ∴ x = 60°
So, ∠ACB = 60° and ∠ACD = 120°
ΔABC is isosceles
...

Example 3:
Prove that in an isosceles triangle, the angle bisector of the apex angle is the
perpendicular bisector of the base
...

To Prove: AD is perpendicular bisector of BC and BD = DC
...

Proof:

In ΔABD and ΔACD,
AB = AC
(Given)
AD = AD
∠BAD = ∠CAD

(Common side)
(AD is a bisector of ∠A)

So, by SAS congruence criterion,
Δ ADB Δ ADC
⇒BD = DC and ∠ADB =∠ADC (CPCT)
Now,
∠ ADB + ∠ADC= 180° (Linear Pair)
⇒∠ ADB + ∠ ADB = 180°
⇒ 2∠ ADB = 180°

Thus, AD is the perpendicular bisector of BC
...
Also, ∠BDC measures 70° and ∠ABC measures 90°
...

∴ ∠BCD = ∠BDC = 70º (∵ Angles opposite equal sides are equal)
In ΔBCD, by the angle sum property, we obtain:
∠BCD + ∠BDC + ∠CBD = 180º
⇒ 70º + 70º + ∠CBD = 180º
⇒ 140º + ∠CBD = 180º
⇒ ∠CBD = 40º
It is given that ∠ABC = 90º
...
Side SU is extended up to point V
such that ST = SV
...


Solution:
In ΔPQR, we have:

PQ = PR (Given)
∴ ∠PQR = ∠PRQ (∵ Angles opposite equal sides are equal)
It is given that ∠QTS = 90°
...

Using the angle sum property in ΔRUS, we obtain:
∠SRU + ∠RUS + ∠RSU = 180°
⇒ ∠SRU + 90° + ∠RSU = 180°
⇒ ∠RSU = 90° ∠SRU … (2)
∠SRU = ∠PRQ (Vertically opposite angles)
Also, ∠PQR = ∠PRQ
∴ ∠SRU = ∠PQR … (3)
Now, from equations (2) and (3), we get:
∠RSU = 90° ∠PQR
⇒ ∠RSU = 90° ∠TQS … (4) [∵ ∠PQR = ∠TQS]
From equations (1) and (4), we get:
∠TSQ = ∠RSU
Hence, QS bisects ∠TSU
...


QS = QS (Common side)
∠TSQ = ∠QSV (∵ ∠TSQ = ∠RSU and ∠RSU = ∠QSV)
ST = SV (Given)
Thus, by the SAS congruency rule, we get:
ΔQTS ≅ ΔQVS

Sides Opposite to Equal Angles of a Triangle are Equal
Observing the Equal Angles and the Sides Opposite to Them in an Isosceles
Triangle
Consider the following ΔPQR
...
However, in ΔPQR, two angles are equal (or congruent)
...
Is the converse of this property also true?
In this lesson, we will study about the equality of sides opposite equal angles in an
isosceles triangle
...

Sides Opposite To Equal Angles of a Triangle Are Equal
Whiz Kid
In an isosceles triangle, the medians drawn from the base vertices to the opposite sides
are of equal length
...
BD and CE are the respective
medians from vertices B and C to sides AC and AB
...

Solved Examples
Easy
Example 1:
In a ΔABC, ∠BAC = 2x and ∠ABC = ∠ACB = x
...

Solution:
It is given that ∠BAC = 2x and ∠ABC = ∠ACB = x
...


We know that the sides opposite equal angles of a triangle are equal
...

Example 2:
In the given ΔABC, ∠ABC = ∠ACB and the perimeter is 11 cm
...


Solution:
It is given that ∠ABC = ∠ACB and AB = 3 cm
...

∴ AC = AB = 3 cm
Perimeter of ΔABC = AB + BC + AC
⇒ 11 cm = 3 cm + BC + 3 cm
⇒ BC = 5 cm

Thus, the length of the base of ΔABC is 5 cm
...
Prove that AD
bisects ∠BAC
...

Example 2:
In the given ΔABC, ∠ABC = ∠ACB
...
Prove that ∠BCD is a right angle
...

Hard
Example 1:
The shown ΔPQR is isosceles with PQ = PR
...
Prove that the medians have the same
length
...

Also, QL and RM are medians
...

Example 2:
In a right-angled triangle ABC, ∠ACB = 2∠BAC
...

Solution:
The given ΔABC can be drawn as is shown
...
Join point A to point D
...
Then, ∠DAB will also be x
...
Their house is located at position B, as shown in the
following figure
...
Rohit chooses the path
AB to reach his house while Mohit first goes to position C starting from A and then
travels distance CB to reach his house
...
Who will
take less time to reach the house?
Since their speeds are the same, more time will be taken by the person who has to
cover the larger distance
...

We can see that the paths followed by Rohit and Mohit form a triangle
...
e
...
From the above example, we can clearly see that length AB is
shorter than length BC + CA
...

Therefore, we can conclude that
“The sum of the lengths of any two sides of a triangle is always greater than the
length of the third side of the triangle”
...

Let us go through the proof of the above property
...


In ΔACD, we have
AD = CA
⇒ ∠ACD = ∠ADC
⇒ ∠ACD + ∠ACB > ∠ADC
⇒ ∠BCD > ∠BDC
⇒ BD > BC

(∠BDC is same as ∠ADC )
(Side opposite to greater angle is longer)

⇒ AB + AD > BC
⇒ AB + CA > BC

(AD = CA)

Similarly, we can prove that AB + BC > CA and CA + BC > AB
...
Similarly, we can
deduce results from the remaining two inequalities
...


The inequalities AB + CA > BC, AB + BC > CA and CA + BC > AB are called triangle
inequalities
...
This tells that the straight line is the shortest path between any two
points
...

Let us now look at some examples to understand this property of triangles better
...

1
...
5 cm, 7
...
7 cm
2
...
4 cm, 18
...
9 cm
Solution:
1
...
5 cm, 7
...
7 cm
For a triangle, the sum of the lengths of any two sides must be greater than the third
side
...
5 cm + 7
...
3 cm > 14
...
5 cm + 14
...
2 cm > 7
...
8 cm + 14
...
5 cm > 11
...

Therefore, a triangle with sides of given lengths is possible
...
7
...
5 cm, 10
...

But in this case,
7
...
9 cm = 18
...
5 cm
Therefore, a triangle with sides of given lengths is not possible
...


Now, we know that the sum of any two sides of a triangle is always greater than the
third side
...

Hence, the sum of the lengths of the four sides cannot be equal to the sum of the
lengths of its two diagonals
...


Which of the following statements is correct?
(i) 2 BD > AB + BC + CA
(ii) 2 BD < AB + BC + CA
Solution:
We know that the sum of any two sides of a triangle is always greater than the third
side
...
e
...


Observing the Relation between the Angles and Sides of a Triangle

Harsh is flying a kite
...
Sometime later, he gives the thread to his friend Mohit who is standing
some distance away from him
...

The figure clearly shows that angle y of ΔACD is greater than angle x of ΔABD
...
This shows us that if we increase the length of any side
of a triangle, then the angle facing that side also increases
...

We will then solve some examples relating to the same
...
Now, what if all the sides of a triangle are unequal? What can be
said about its angles? The triangle inequality theorem describes such a triangle
...


Consider the given ΔABC
...

AB = 8 cm is the longest side in ΔABC
...
e
...
Also, AC = 3 cm is the shortest side in ΔABC
...
e
...

Thus, we can conclude that the angle opposite the longer side is greater
...

Proving the Triangle Inequality Theorem
Given: ΔPQR in which PR > PQ
To prove: ∠Q > ∠R
Construction: Mark a point S on PR such that PQ = PS
...


PP

Proof: In ΔPQS, we have:
PQ = PS (By construction)
⇒ ∠1 = ∠2
...

∴ ∠2 > ∠SRQ

⇒ ∠2 > ∠PRQ … (2) [∵ ∠PRQ = ∠SRQ]
From (1) and (2), we have:
∠1 > ∠PRQ
...

So, ∠PQR > ∠1 … (4)
Thus, from (3) and (4), we can conclude that:
∠PQR > ∠PRQ
Solved Examples
Easy
Example 1:
In the given ΔPQR, PQ is greater than PR
...
Prove that ∠SRQ > ∠SQR
...
Prove that ∠BAC > ∠ABC
...


In ΔABC, we have:
∠ABC = ∠ACB … (1) [∵ Angles opposite equal sides AB and AC are equal]
By the triangle inequality theorem, we have:
∠ACB < ∠BAC … (2) [∵ Angle opposite longer side BC is greater]
Thus, from (1) and (2), we obtain:
∠BAC > ∠ABC
Medium
Example 1:
In the shown ΔABC, AB = 4 cm, BC = 5 cm and the perimeter is 16 cm
...


Solution:
In ΔABC, we have:
AB = 4 cm and BC = 5 cm (Given)
Perimeter = 16 cm (Also given)
⇒ AB + BC + CA = 16 cm (∵ Perimeter is the sum of all sides)
⇒ 4 cm + 5 cm + CA = 16 cm
⇒ CA = 7 cm
Now, CA = 7 cm is the longest side of ΔABC
...
e
...

∴ ∠ABC > ∠BAC and ∠ABC > ∠ACB … (1)
Also, BC > AB
∴ ∠BAC > ∠ACB … (2)
Thus, from (1) and (2), we can conclude that ∠ABC is the greatest angle and ∠ACB is
the smallest angle in ΔABC
...
Then, prove that ∠BCD > ∠BAD
...

So, AB > BC
⇒ ∠1 > ∠2 … (1) [∵ Angle opposite longer side is greater]
In ΔADC, CD is the shortest side
...
Show that ∠PRS > ∠PSR
...

So, ∠PQR = ∠PSQ + ∠SPQ

(By the exterior angle property)

⇒ ∠PQR > ∠PSQ … (2)
From (1) and (2), we obtain:
∠PRQ >∠PSQ
⇒ ∠PRS > ∠PSR (∵ ∠PRQ = ∠PRS and ∠PSQ = ∠PSR)
Example 2:
In a ΔABC, AB > AC and AD is the bisector of ∠BAC
...

Solution:
Let the following ΔABC be the given triangle such that AB > AC
...


In ΔABC, we have:
AB > AC (Given)

⇒ ∠ACB > ∠ABC (∵ Angle opposite longer side is greater)
On adding ∠1 to both sides, we get:
∠ACB + ∠1 > ∠ABC + ∠1
⇒ ∠ACB + ∠2 > ∠ABC + ∠1 … (1) [∵ AD bisects ∠BAC; ∠1 = ∠2]
By the exterior angle property, we have:
∠ADB = ∠ACB + ∠2 … (2)
Similarly, ∠ADC = ∠ABC + ∠1 … (3)
Thus, by using (1), (2) and (3), we can conclude that:
∠ADB > ∠ADC

Observation of the Sides of a Triangle by Seeing the Angles
Consider the following triangular racetrack where two cars start from different points A
and B to reach the finish point C
...
Do you think any one car has an
advantage over the other?
On observing the triangular track, it seems that path BC is longer than path AC
...
Also, the angle opposite BC is
greater than the angle opposite AC
...
We will also solve some problems based on this relation
...
The converse of this property is also true
...
In other words, the smaller angle has the shorter side opposite it
...


Let us apply the stated property in this triangle
...

Therefore, the side opposite ∠QPR, i
...
, QR is the longest side of the triangle
...

Therefore, the side opposite ∠PQR, i
...
, PR is the shortest side of the triangle
...
Using this property, we can say
that in a right triangle, the hypotenuse is the longest side
...


The given triangle is an example of a skinny triangle
...

Proof: In ΔPQR, we can have three possible cases: (1) PQ > PR, (2) PQ = PR and (3)

PQ < PR
...
Thus, PQ > PR is not true
...
Thus, PQ = PR is also not
true
...
e
...

Thus, we have proven that if two angles of a triangle are unequal, then the greater
angle has the longer side opposite it
...
Find the greatest angle and identify the
longest side of the triangle
...

Let the angles be x, 2x and 3x, as is shown in the figure
...
e
...

Now, we know that the side opposite the greater angle is longer
...

Example 2:
Which side of the given triangle is the shortest?

Solution:
In order to find the shortest side of ΔDEF, we need to figure out the smallest angle of
the triangle
...

By the angle sum property of triangles, we have:
∠EDF + ∠DEF + ∠EFD = 180°
⇒ 48° + 87° + ∠EFD = 180°
⇒ 135° + ∠EFD = 180°
⇒ ∠EFD = 45°
Clearly, ∠EFD has the smallest measure in the given triangle
...
e
...

Medium
Example 1:
In the given ΔABC, ∠ABD > ∠ACD and AD is the bisector of ∠BAC
...
AB < AC
2
...
It is given that ∠ABD > ∠ACD
...
In ΔABD, we have:
∠ABD + ∠BAD + ∠ADB = 180° (By the angle sum property)
⇒ ∠ACD + ∠BAD + ∠ADB < 180° (∵ ∠ABD > ∠ACD)
⇒ ∠ACD + ∠CAD + ∠ADB < 180° (∵ AD bisects ∠BAC; ∠BAD = ∠CAD)
⇒ ∠ADB < 180° − (∠ACD +∠CAD)
⇒ ∠ADB < ∠ADC (By the angle sum property in ΔADC)
Example 2:
Prove that of all the line segments that can be drawn to a given line from a point lying
outside the line, the perpendicular line segment is the shortest
...
Also, PM ⊥ m and N is any point other than M on m
...

In ΔPMN, we have:
⇒ ∠PMN + ∠MPN + ∠PNM = 180° (By the angle sum property)
⇒ 90° + ∠MPN + ∠PNM = 180°
⇒ ∠MPN + ∠PNM = 90°
⇒ ∠PNM < 90°

⇒ ∠PNM < ∠PMN
⇒ PM < PN (∵ Side opposite greater angle is longer)
Thus, PM is the shortest of all line segments from point P to line m
...
Prove that QS > SR
...


So, ∠PSQ + ∠PSR = 180°
⇒ 30° + ∠PSR = 180°
⇒ ∠PSR = 180° − 30°
⇒ ∠PSR = 150°
In ΔPSR, we have:
∠PSR + ∠SPR + ∠PRQ = 180° (By the angle sum property)
⇒ 150° + ∠SPR + 20° = 180°
⇒ ∠SPR = 180° − (150° + 20°)
⇒ ∠SPR = 10°
So, ∠PSR > ∠PRQ > ∠SPR
Since the side opposite the greater angle is longer, we get:
PR > PS > SR … (2)
By using (1) and (2), we can conclude that:
QS > SR
Example 2:
Prove that the difference between any two sides of a triangle is less than the third side
...
We will do so by showing that in ΔABC,
BC − AC < AB
...
Join C to D
...

∴ ∠ACD = ∠ADC
⇒ ∠ACD + ∠ACB > ∠ADC
⇒ ∠BCD > ∠ADC
We know that the side opposite the greater angle is longer
...

Thus, we have proved that the difference between any two sides of a triangle is less
than the third side

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