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Geometry

Lines & Triangles

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CAT 2025 Lesson : Lines & Triangles - Other Theorems

3.8 Other Theorems

Definition & Features	Figure
Apollonius Theorem: In a triangle ABC, where AD is the median, AB $^2$ + AC $^2$ $=$ 2(AD $^2$ + BD $^2$ )
Extension of Apollonius: In a triangle ABC, where AD, BE and CF are the medians, 4 (AD $^2$ + BE $^2$ + CF $^2$ ) = 3 (AB $^2$ + BC $^2$ + AC $^2$ )
Interior Angle Bisector Theorem: In a triangle ABC, where AD is the interior angle bisector, $\mathrm {\dfrac{AB}{AC} = \dfrac{BD}{CD}}$
Exterior Angle Bisector Theorem: In a triangle ABC, where AD is the external angle bisector of $\angle$ A and D is a point on extended BC, $\mathrm {\dfrac{AB}{AC} = \dfrac{BD}{CD}}$
Basic Proportionality Theorem: A line drawn parallel to one of the sides cuts the other 2 sides in the same ratio. In $\triangle$ ABC, DE \| \| BC. ∴ $\mathrm {\dfrac{AD}{DB} = \dfrac{AE}{EC}}$
Midpoint Theorem: Extending the above theorem, we note that the line joining the mid points of any 2 sides of a triangle will be parallel to the third side. In the triangle ABC, where D and E are the midpoints of AB and AC, DE \| \| BC. Likewise, if a line that bisects one side of the triangle is parallel to another side, then it definitely bisects the third side.

Example 19

\triangle

ABC, E is the mid-point of AC and D is a point on AB. The bisector of

\angle

A intersects DE and BC at P and Q respectively. QC

=

3

DP and BQ

=

PE. If DP : PE

=

2 : 3

, then AD : DB

=

Solution

As AQ is the angle bisector of

\angle

A, using angle bisector theorem,

The lengths of these sides are denoted with

x

and

y

=

3

=

6x

As E is the midpoint, AE

=

=

3y

\triangle

AQC,

\dfrac{AE}{AC} = \dfrac{PE}{QC} = \dfrac{1}{2}

\therefore

Using inverse of the Basic Proportionality Theorem, we note that PE | | QC, which implies DE | | BC.

The line joining the midpoint of AC and parallel to BC, will pass through the midpoint of AB (Midpoint Theorem).

\therefore

D is the mid-point of AB and AD : DB

=

2 : 1

Answer:

2 : 1

Example 20

\triangle

ABC AB

=

5

cm, AC

=

7

cm and BC

=

4

cm. D is a point on BC such that BD

=

1

cm. Then AD

=

?

(1)

\sqrt{14} \space \space \space\space \space \space \space\space \space \space

(2)

2\sqrt{7} \space \space \space\space \space\space \space \space \space \space

(3)

\sqrt{20} \space \space \space\space \space\space \space \space \space \space

(4)

2\sqrt{10} \space \space \space

Solution

Let E be the mid-point of BC making AE the median. In

\triangle

ABC, using Apollonius theorem,
AB

^{2}

+

^{2}

=

2

(AE

^{2}

+

^{2}

)

5^{2} + 7^{2} =

2

( AE

^{2}

2^{2})

⇒ AE

=

\sqrt{33}

\triangle

ABE, using Apollonius theorem,
AB

^{2}

+

^{2}

=

2

(AD

^{2}

+

^{2}

)

5^{2} + (\sqrt{33})^{2} = 2(

^{2} + 1^{2})

⇒ AD

=

\sqrt{28} = 2\sqrt{7}

Answer: (2)

2\sqrt{7}

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