HELP!! How to solve these circle geom questions ! (1 Viewer)

[kasser]

1)Two circles touch internally at A. The tangent at P on the smaller circle cuts the larger circle at Q and R, prove AP bisects angle RAQ?

2)PA and PB are two tangents to a circle and x is the midpoint of minor arc AB.Prove XA bisects angle PAB and XB bisects angle PBA?

3)AB and CD are parallel chords of a circle. The tangent at B meets CD produced at E. Prove triangle ABC and triangle DBE are similar.

4)Two circles intersect at A and B. A line through A cuts the first circle at P and second circle at Q. From an external point T, a tangent TP is drawn and TQ produced meets the second circle again at R Prove that the points P, T, R and B are con cyclic ?

These were the questions i couldn't do out of the 4 chapters.

 

[TL1998]

Okay I was hoping someone would answer your questions so i just ignored this thread but no body answered so......

Question 1) Once you construct a tangant at the point A, you just need to alt segment theorem and the angle sum of a triangle to figure this out. If you still need help, PM me

Question 2) Since X is the midpoint of AB, what does this tell you?? That's right, the straight lines XB and XA are equal, meaning that triangle BXA is isosceles. Then you use alt segment theorem, which shows that angle XAP equals to angle XBA, but since triangle BXA is isoscles, angle XBA = angle angle BAX. Thus angle BAX = angle XAP, so XA bisects the angle

Question 3) If you join A to C, and B to D, you'll be able to see that ABCD is a cyclic quad. so angle BDE = angle BAC (since exterior angle of a cyclic quadrilateral equals the opposite interior angle). Also, the angle DBE equals angle BCD, (alt seg theorem). But since AB is parallel to CD, angle BCD = angle CBA. since two angles are the same, the two triangles are equal.

Question 4) NOTE: I labelled a point opposite to T on the line PT produced as 'X'. Let angle XPB = angle alpha. then by alt seg, angle PAB is also alpha. Then angle BAQ = 180 - alpha (straight line angle is 180). But angle BAQ = Angle BRQ (angles in the same segment are equal) = 180-alpha. But angle BRQ + angle TRB = 180. So angle TRB = alpha. Since angle XPB = angle TRB, PTRB are concyclic. (exterior angle is equal to opposite interior angle)

 

[kasser]

Thankyou, this actually helped, i think i learned something new