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Exterior Angle Property

Exterior Angle: If the side BC of a triangle ABC is produced to form ray BD, then $large angle ACD$ is called an exterior angle of $large Delta ABC$ at C and is denoted by ext. $large Delta ACD.$

Theorem: If a side of a triangle is produced, the exterior angle so formed is equal to the sum of the two interior opposite angles.

Given: A triangle ABC. D is a point on BC produced, forming an exterior angle $large angle 4.$

To Prove: $large angle 4=angle 1+angle 2$

i.e., $large angle ACD=angle CAB+angle CBA.$

Proof: In triangle ABC, we have

$large angle 1+angle 2+angle 3=180^{circ}$ ...........(i)

Also, $large angle 3+angle 4=180^{circ}$ [ $large because angle 3$ and $large angle 4$ forms a linear pair] ...(ii)

Find (i) and (ii), we have

$large angle 1+angle 2+angle 3=angle 3+angle 4$

$large Rightarrow;;;angle 1+angle 2=angle 4$

Hence, $large angle 4=angle 1+angle 2$

i.e. $large angle ACD=angle CAB+angle CBA$

Illustration: in the figure PQ || RS. Find angle QOR.

Solution: Extend PQ to N

$angle PNR = angle SRN= 120^0$

[ Alternate interior angles are equal PN || RS and RO is the transversal ]

$angle PNR + angle QNO= 180^0$ [ Linear Pair]

$120^0 + angle QNO= 180^0$

$Rightarrow angle QNO= 180^0-120^0=60^0$

Now, In $bigtriangleup NOQ$ , we have

$angle PQO= angle QNO + angle NOQ$ [ Exterior angle Property]

$100^0= angle 60^0 + angle NOQ$

$angle NOQ= 100^0-60^0= 40^0$

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Exterior Angle Property