difficult circle geometry questions :( (1 Viewer)
- Thread starter norelle
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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
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
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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
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
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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
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Where are these questions from ?cant do these 3 questions..
please help~~thankss
they are from catholic trialsWhere are these questions from ?