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Lines & Triangles

Lines And Triangles

MODULES

Lines & Angles
Parallel Lines
Basics of Triangles
Types of Triangles
Triangle & its Segments
Area of a Triangle
Isosceles & Equilateral
Right-angled Triangles
Other Theorems
Congruency & Similarity
Past Questions

CONCEPTS & CHEATSHEET

Concept Revision Video

SPEED CONCEPTS

Lines & Triangles 1
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Lines & Triangles 2
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Lines & Triangles 3
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PRACTICE

Lines & Triangles : Level 1
Lines & Triangles : Level 2
Lines & Triangles : Level 3
ALL MODULES

CAT 2025 Lesson : Lines & Triangles - Other Theorems

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3.8 Other Theorems

Definition & Features Figure
Apollonius Theorem: In a triangle ABC, where AD is the median,

AB2^22 + AC2^22 === 2(AD2^22 + BD2^22)
Extension of Apollonius: In a triangle ABC, where AD, BE and CF are the medians,

4 (AD2^22 + BE2^22 + CF2^22) = 3 (AB2^22 + BC2^22 + AC2^22)
Interior Angle Bisector Theorem: In a triangle ABC, where AD is the interior angle bisector,

ABAC=BDCD\mathrm {\dfrac{AB}{AC} = \dfrac{BD}{CD}}ACAB​=CDBD​
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,

ABAC=BDCD\mathrm {\dfrac{AB}{AC} = \dfrac{BD}{CD}}ACAB​=CDBD​
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.
∴ ADDB=AEEC\mathrm {\dfrac{AD}{DB} = \dfrac{AE}{EC}}DBAD​=ECAE​
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

In △\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 === 333DP and BQ === PE. If DP : PE === 2:32 : 32:3, 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 xxx and yyy.
=== 333DP === 6x6x6x
As E is the midpoint, AE === EC === 3y3y3y
In △\triangle△AQC, AEAC=PEQC=12\dfrac{AE}{AC} = \dfrac{PE}{QC} = \dfrac{1}{2}ACAE​=QCPE​=21​


∴\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 === 2:12 : 12:1

Answer: 2:12 : 12:1


Example 20

In △\triangle△ABC AB === 555cm, AC === 777cm and BC === 444cm. D is a point on BC such that BD === 111cm. Then AD === ?

(1) 14          \sqrt{14} \space \space \space\space \space \space \space\space \space \space14​           (2) 27          2\sqrt{7} \space \space \space\space \space\space \space \space \space \space27​           (3) 20          \sqrt{20} \space \space \space\space \space\space \space \space \space \space20​           (4) 210   2\sqrt{10} \space \space \space210​   

Solution

Let E be the mid-point of BC making AE the median. In △\triangle△ABC, using Apollonius theorem,
AB2^{2}2 +++ AC2^{2}2 === 222(AE2^{2}2 +++ BE2^{2}2)
52+72=5^{2} + 7^{2} =52+72= 222( AE2^{2} 2+22) 2^{2})22)
⇒ AE === 33\sqrt{33}33​

In △\triangle△ABE, using Apollonius theorem,
AB2^{2}2 +++ AE2^{2}2 === 222(AD2^{2}2 +++ BD2^{2}2)
52+(33)2=2(5^{2} + (\sqrt{33})^{2} = 2(52+(33​)2=2( AD2+12)^{2} + 1^{2})2+12)

⇒ AD === 28=27\sqrt{28} = 2\sqrt{7}28​=27​

Answer: (2) 272\sqrt{7}27​


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