**Answer
1** :
In the given problem, the exterior angles obtained on producing the base of a triangle both ways are and . So, let us draw ΔABC and extend the base BC, such that:

Here, we need to find all the three angles of the triangle.

Now, since BCD is a straight line, using the property, “angles forming a linear pair are supplementary”, we get

Similarly, EBC is a straight line, so we get,

Further, using angle sum property in ΔABC

Therefore,

**Answer
2** :

In the given problem, *BP *and *CP *arethe internal bisectors of respectively. Also, *BQ and CQ arethe external bisectors of respectively. Here, we need toprove:*

*We know that if the bisectors of anglesand of Δ ABC meetat a point O then *

*Thus, in Δ ABC*

* ……(1)*

Thus, Δ*ABC*

∠BQC=90°−12∠A ......(2)∠BQC=90°-12∠A ......2

Adding (1) and (2), we get

Thus,

Hence proved.

**Answer
3** :

In the given Δ*ABC*, and . We need to find .

Here, are vertically oppositeangles. So, using the property, “vertically opposite angles are equal”, we get,

Further, *BCD* is a straight line. So, usinglinear pair property, we get,

Now, in Δ*ABC*, using “the angle sum property”, we get,

Therefore,

(i)

(ii)

(iii)

(iv)

**Answer
4** :

In the given problem, we need to find the value of *x*

(i) In the given Δ*ABC*, and

Now, *BCD* is a straight line. So, using theproperty, “the angles forming a linear pair are supplementary”, we get,

Similarly, *EAC* is a straight line. So, we get,

Further, using the angle sum property of a triangle,

In Δ*ABC*

Therefore,

(ii) In the given Δ*ABC*, and

Here, *BCD *is a straight line. So, using theproperty, “the angles forming a linear pair are supplementary” we get,

Similarly, *EBC *is a straight line. So, we get

Further, using the angle sum property of a triangle,

In Δ*ABC*

Therefore,

(iii) In the given figure,and

Here,and *AD is thetransversal, so form a pair of alternate interiorangles. Therefore, using the property, “alternate interior angles are equal”,we get,*

*Further, applying angle sum property of the triangle*

*In Δ DEC*

*Therefore, *

*(iv) In the given figure,,and *

*Here, we will produce AD to meet BC at E*

*Now, using angle sum property of the triangle*

*In Δ AEB*

*Further, BEC is a straight line. So, using theproperty, “the angles forming a linear pair are supplementary”, we get,*

*Also, using the property, “an exterior angle of a triangle isequal to the sum of its two opposite interior angles”*

*In Δ DEC, x is its exterior angle*

*Thus,Therefore,.*

In the given figure, AB divides ∠DAC in the ratio 1 : 3 and AB = DB. Determinethe value of x.

**Answer
5** :

In the given figure,and

Since,and angles opposite to equal sidesare equal. We get,

∠BDA=∠BAD .....(1)∠BDA=∠BAD .....1

Also, *EAD *is a straight line. So, using theproperty, “the angles forming a linear pair are supplementary”, we get,

Further, it is given *AB* divides in the ratio 1 : 3.

So, let

∠DAB=y, ∠BAC=3y∠DAB=y, ∠BAC=3y

Thus,

y+3y=∠DAC⇒4y=72°⇒y=72°4⇒y=18°y+3y=∠DAC⇒4y=72°⇒y=72°4⇒y=18°

Hence, ∠DAB=18°, ∠BAC=3×18°=54°∠DAB=18°, ∠BAC=3×18°=54°

Using (1)

Now, in Δ*ABC* , using the property, “exterior angleof a triangle is equal to the sum of its two opposite interior angles”, we get,

∠EAC=∠ADC+x⇒108°=18°+x⇒x=90°∠EAC=∠ADC+x⇒108°=18°+x⇒x=90°