difficult circle geometry questions :( (1 Viewer)

[norelle]

cant do these 3 questions..
please help~~thankss

 

[Drongoski]

Q1

i) since M is mid-pt of CB, it is centre of semi-circle passing thru X, C and B (since angle BXC is right-angled); MX is therefore radius of this semi-circle. Thus MB = MX so that angle MBX = angle MXB, being the base angles of isosceles triangle MBX.

ii) angle DXN = angle MXB = @ say (vert opp angles). .: angle MBX = angle CBD (same angle) = @ = angle CAD (angles in same segment, on a common chord CD)

In triangles DXA and DNX:

angle XAD = angle NXD (shown)
angle XDA = angle NDX (same angle)
.: the two triangles are equiangular and therefore similar.
.: angle XND = corresponding angle AXD = 90 deg
.: MN perpendicular to AD

 

[Drongoski]

Q2

Join the line MN

Let angle BAD = 2 @ ==> angle NAB = @

Let angle DCB = 2 # ==> angle MCB = #

.: @ + # = 90 deg (opp angles BAD + DCB = 180 deg in a cyclic quadrilateral)

Now angle MNB = angle MCB = # (angles on common arc in same segment)
similarlly: angle BMN = angle BAN = @

.: for triangle MNB, angle MNB + angle BMN = # + @ = 90 deg

.: the remaining angle MBN = 90 deg ( a right angle)

.: MN is diameter of semicircle thru M, B and N

But this semicircle is part of the original circle

.: it is diameter of the original circle

 

[Drongoski]

cant do these 3 questions..
please help~~thankss

Q3

I found this question hard initially !


i) since MN is perp to GD angle GNM = 90 deg
and angle BMN = angle GNM = 90 deg (Alt angles, AB//GD)
.: MN perpendicular to AB and since it bisects chord AB, MN passes thru the centre of the circle
.: it also bisects GD
.: GN = ND
.: triangle MNG congruent to triangle MND (can be shown: SAS)
.: base angles MGN = MDN = @ say.

ii) angle BMD = angle MDN = @ (alt angles, AB//GD)
similarly angle AMG = angle MGN = @
also angle AMC = angle BMD = @ (vert opp angles)
and angle BPR = angle GRP (alt angles, AB//GD) (R is where CF intersects GD)
Now angle FGD = angle FCD = # say (both on same chord in same segment)
But angle BPR = @ + # (ext angle of triangle = sum of int opp angles)
.: angle GRP = @ + # = angle RGF + angle RFG (ext angle of triangle . . .)
.: angle RFG = @ since angle RGF = #

Therefore angle PMG = angle PFG
But they are both subtended by the same chord PG
therefore PMFG is a cyclic quadrilateral

iii) .: angle PGM = angle PFM (on same chord PM and same segment; PMFG cyclic)
Also angle CFM (i.e. CFE) = angle CDE (same chord, same segment)

In triangles PGM and QDM:

angle PMG = angle QMD (shown earlier)
angle PGM = angle QDM (shown)
MG = MD (equal sides of isosceles triangle MGD)
.: triangles PGM and QDM are congruent (AAS)
.: corresponding sides PM = MQ

QED

 

[Drongoski]

cant do these 3 questions..
please help~~thankss

Where are these questions from ?

 

[norelle]

Where are these questions from ?

they are from catholic trials

 

[norelle]

thankss for helping with the questions! =D