Anyone That Can Please Help me with Circle Geo Questions? (1 Viewer)

[181jsmith]

See 2 questions below, thanks

1. Attachment 24581

In the diagram, the bisector of the

meets RQ in S and the circum-circle of the triangle PQR in T.

Prove that

2. Attachment 24582

AB is the common chord of two circles C1 and C2. AC and AD are chords of the respective circles with

and CD meets the circles at P and Q respectively. R is the foot of the perpendicular from P to BQ. Prove that

 

[Sanjeet]

1. The two triangles are congruent (top angles are the same, PS is common and PSQ=PSR=90, or equiangular)
Thus through pythogoras' theorom PS^2 = PQ^2 - SQ^2. But SQ = RS and PQ = PR (as they are congruent)
Therefore PS^2 = PQ x OE -RS x SQ
I'm at tafe right now hence I can't use the proper maths syntax, and I can't be botherd to go through Q2 right now so I'll look at it when I get home

 

[Aesytic]

2. In C1,
angle CAB = angle CPB (angles standing on the same arc are equal; arc BC)
In C2,
angle DAB = angle DQB (angles standing on the same arc are equal; arc BD)
Since angle CAB = angle DAB,
.'. angle CPB = angle DQB
If we let angle CPB and angle DQB be x,
In triangle PBQ,
angle PBQ = 180 - x - x (angle sum of a triangle is 180)
= 180 - 2x
Also, in triangle PRQ,
angle PRQ = 90 (given)
angle DQB = x (proven)
.'. angle RPQ = 180 - 90 - x (angle sum of a triangle is 180)
= 90 - x
Looking at before,
angle PBQ = 180 - 2x
= 2*(90-x)
=2*angle RPQ