Problem – 1: In the **∆**ABC, if the bisectors of ∠B and ∠C meet the point P and their exterior bisectors at Q, show that the four points B, P, C and Q are concyclic.

General Enunciation: In the ABC, if the bisectors of **∠**B and **∠**C meet the point P and their exterior bisectors at Q, We have to show that the four points B, P, C and Q are concyclic.

Particular Enunciation: Suppose in the **∆**ABC, BP and CP are the interior bisector of **∠**B, **∠**C respectively and they meet at the point P. Whether BQ and CQ are exterior bisectors of those two angles respectively and they meet at the point Q. let us Show that the four points B,P,C, and Q are concyclic.

Proof: In quadrilateral BPCQ we get

**∠**PBQ+**∠**PCQ = ½ **∠**PBC+ ½ **∠**QBC+ ½ **∠**PCB+ ½ **∠**QCB

= ½ **∠**B+ ½ **∠**CBE+ ½ **∠**C+ ½ **∠**BCF

= ½ **∠**B+ ½ (**∠**A+**∠**C) + ½ (**∠**A+**∠**B)

= ½ (2**∠**A+2**∠**B+2**∠**C) =**∠**A+**∠**B+**∠**C

i.e., **∠**PBQ+**∠**PCQ =**∠**A+**∠**B+**∠**C

But in **∆**ABC = **∠**A+**∠**B+**∠**C= 180^{0}

∴ **∠**PBQ+**∠**PCQ=**∠**A+**∠**B+**∠**C=180^{0}

Since the sum opposite angles of the quadrilateral BPCQ is 180^{0}, hence the four points B, P, C and Q are concyclic. __(Showed__) ** **

Problem – 2: Prove that the bisector of any angle of cyclic quadrilateral and the exterior bisector of its opposite angle meet one the circumference of the circle.

General Enunciation: we have to prove that the bisector of any angle of cyclic quadrilateral and the exterior bisector of its opposite angle meet one the circumference of the circle.

Particular Enunciation: Suppose, ABCD be the cyclic quadrilateral. BE is the bisector of the interior angle **∠**B and E. Let us prove that E is one the circumference.

Proof: Since ABCD is the cyclic quadrilateral then we get

**∠**ABC+**∠**ADC =two right angle ———————- (1)

Again **∠**ADC +**∠**CDF =one straight angle ———–(2)

From equation (1) and (2) we get **∠**ABC + **∠**ADC =<ADC+**∠**CDF

∴ **∠**ABC =**∠**CDF

∴ ½ **∠**ABC= ½ **∠**CDF=**∠**CDF

Now in quadrilateral ABED we get

**∠**ABE+**∠**ADC +**∠**CDF= ½ **∠**ABC =+**∠**ADC + ½ **∠**ABC

**∠**ABC =

**∠**ABE and <CDF = ½

**∠**ABC]

∴ **∠**ABE + **∠**ADC +**∠**CDF=**∠**ADC +<ABC=180

∴ The quadrilateral ABED is cyclic E on the circumference of the circle __(Proved__).

Problem – 3: ABCD is a circle. If the bisectors of ∠CAB ∠CBA meet at the point P and the bisectors of ∠DBA and ∠DAB meet at Q, prove that the four point A, Q, P and B are concyclic.

General Enunciation: ABCD is a circle. If the bisectors of **∠**CAB **∠**CBA meet at the point P and the bisectors of **∠**DBA and **∠**DAB meet at Q, we have prove that the four point A, Q, P and B are concyclic**. **

**Particular enunciation**: Given that ABCD is a circle. The bisectors of **∠**CAB and **∠**CBA are AP and BP respectively meet at the point P and the bisectors of **∠**DBA and **∠**DAB are BQ and AQ respectively meet at Q. let us prove that the four points A,Q,P and B are concyclic.

**Proof: **In **∆**ABC, **∠**A+**∠**B+**∠**C=180^{0} ——————- (1)

In **∆**PAB, **∠**P+**∠**PAB+**∠**PBA=180^{0}

Or ,**∠**P+ ½ **∠**A+ ½ **∠**B=180^{0}

Or, **∠**P+ ½ **∠**A+ ½ **∠**B +½ **∠**C=180^{0}+ ½ **∠**C

Or, **∠**P+ ½ (**∠**A+**∠**B+**∠**C)= 180^{0} ½ +**∠**C

Or **∠**P + ½×180^{0}=180^{0}+ ½ **∠**C

Or, **∠**P+90^{0}=180^{0}+ ½ **∠**C

Or, **∠**P=90+ ½ **∠**C ———————– (2)

Similarly in **∆**ABD we can prove that **∠**Q=90+ ½ **∠**C ———– (3)

Then from equation (2) and (3) we get

**∠**P =**∠**Q=90+ ½ **∠**C

i.e., **∠**P=**∠**Q

Now AB is the arc of the circle and **∠**P and **∠**Q are circular angles and **∠**P= **∠**Q.

∴ A, Q, P and B are the concyclic [since the two circular angles are equal] **(**Proved)** **