The bisector of ZB of an isosceles AABC with AB ACmeets the circumcirc

1.	The bisector of ZB of an isosceles AABC with AB ACmeets the circumcircle of AABC at P. If AP and BCproduced meet at Q, prove that : CQ = CA
Answer» Explanation:-Since AB = AC, hence angle opposite to equal sides are also equal. < ACB = < ABC. Since, ext < ACB = < QAC + < AQC Then, < ABC = < QAC + < AQC Now, because < ABP = < PBC. Therefore, 2 < PBC = < ABC Hence, 2 < PBC = < QAC + < AQC Now, < QAC = < PAC and because ∆PAC and ∆PBC lie on same line segment PC. So, < PAC = < PBC Hence < PBC = < QAC 2 < PBC = < PBC + AQC < PBC = < AQC < PAC = < AQC < QAC = < AQC Side opposite to equal sides are equal. QC = AC Hence prove.

The bisector of ZB of an isosceles AABC with AB ACmeets the circumcircle of AABC at P. If AP and BCproduced meet at Q, prove that : CQ = CA

Answer»

Explanation:-Since AB = AC, hence angle opposite to equal sides are also equal.

< ACB = < ABC.

Since, ext < ACB = < QAC + < AQC

Then, < ABC = < QAC + < AQC

Now, because < ABP = < PBC.

Therefore, 2 < PBC = < ABC

Hence, 2 < PBC = < QAC + < AQC

Now, < QAC = < PAC and because ∆PAC and ∆PBC lie on same line segment PC.

So, < PAC = < PBC

Hence < PBC = < QAC

2 < PBC = < PBC + AQC

< PBC = < AQC

< PAC = < AQC

< QAC = < AQC

Side opposite to equal sides are equal.

QC = AC

Hence prove.