O lvl A Maths: Plane Geometry

Question from http://www.sgforums.com/forums/2297/topics/368889

In the diagram, XTPY is a tangent to 2 circles at P and ABTC is a tangent to the circle of centre O at C.
The line AP is a chord of one circle and produced to meet the other circle at Q.

(i) Explain why a circle passes through O,C, T and P

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Answer:

Let angle COP = 2x

Hence angle CQP=x (angle at centre = 2 angle at circumference)

angle PCB = angle CQP =x (alt. segment theorem)

Also, angle CPT= angle CQP=x (alt. segment theorem)

Hence, angle CTP= 180 - 2x (sum of angles in triangle)

Since angle COP + angle CTP = 180, quadrilateral CTPO is a cyclic quad, at which a circle will pass through all of the points. (shown)


Thanks to Leekeewei at sgforums for providing the solution.

O lvl A Maths: Plane Geometry

Question from http://www.sgforums.com/forums/2297/topics/368877

In the diagram, P is any point on the semicircle centre O, and PQ is perpendicular to AB. The inscribed circle centre C touches PQ, AB and the semicircle at D, E and F respectively. Prove that

(a) A,D and F lie on the same straight line

(b) AD * AF = AQ * AB

(c) AE² = AQ² + AQ * QB

(d) AE = AP

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Answer:


(a) DC is perpendicular to PQ, hence DC is parallel to AO.
OCF is a straight line.

=> angle DCF = angle AOF
=> FDC and FAO are similar triangles.

Hence, A, D and F lies on the same straight line.


(b) Draw a straight line from F to B.

angle AFB = 90 degrees = angle AQD
angle DAQ = angle BAF

=> angle ADQ = angle ABF
=> ADQ and ABF are similar triangles.
=> AD / AB = AQ / AF

AD X AF = AQ X AB (proved)


(c)
AE² = AF * AD( tangent-secant theorem)

From part (b) AD X AF = AQ X AB
AE² = AQ * AB
AE² = AQ ( AQ + AB)
AE² = AQ² + AQ * AB (proven)


(d) Draw a full circle and extend a line to form a chord


AD = AQ * QD
PQ = QZ ( radius from centre bisect chord)
PQ * QZ = PQ²

Therefore, PQ² = AQ * QB ( Intersecting chords theorem)

From part (c) : AQ² = AE² - (AQ)(QB)
AP² = AQ² + PQ²

Therefore, AP² = [AE² - (AQ)(QB)] + [(AQ)(QB)] ( Replaced found eqn)
AP² = AE²
AP = AE (proved)




Thanks to forumers at sgforums for providing the question and answer :D

O lvl A Maths: Plane Geometry

Question from http://www.sgforums.com/forums/2297/topics/359741

A line PQ is drawn through the verex A of triangle ABC such that BP and CQ are perpendicular to PQ. If M is the midpoint of BC and MR is perpendicular to PQ, prove that MP = MQ.

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Answer:

We can draw an auxiliary line (in red) to help us see easier.
Draw it such that CH is perpendicular to MR and CG perpendicular to BP
(Note: BP is parallel to MR and parallel to CQ since all are perpendicular to PQ)

We can see that triangle BCG and MCH are similar, with BM = MC
Hence, by intercept theorem, GH = HC

Since PGHR and RHCQ are rectangles, PR = GH = HC = RQ
PR = RQ

Since PR = RQ and MR is perpendicular to PQ, then triangle MPQ is isosceles.
Hence, MP = MQ (shown)

O lvl A Maths: Plane Geometry

Question from http://www.sgforums.com/forums/2297/topics/361016

In the trapezium ABCD, AD is parallel to BC, BC = 3AD and the diagonals AC and BD meet at E. The line through A is drawn parallel to DB to meet the extended line of CB at F. Prove that

(a) FB = AD
(b) EC = 3AE

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Answer:

(a) Since AF is parallel to BD, and AD is parallel to BC (because F is an extension of BC, hence making both FB and FC also parallel to AD ), ADBF must be a parallelogram, with FB = AD.

(Extra notes: The opposite sides of a parallelogram are always of equal lengths.)

(b) Since BC = 3AD, and AD is parallel to BC, triangle AED and triangle BEC must be similar triangles. As triangle AED and triangle BEC are similar triangles, all sides of triangle BEC must be 3 times the similar sides of triangle AED, hence EC = 3AE.


P.S. Credits to ForbiddenSinner for the solutions and explanations.

O lvl A Maths: Plane Geometry

Question from http://www.sgforums.com/forums/2297/topics/360786

The diagonals of cyclic quadrilateral PQRS intersect at U. The circle's tangent at R meets PS produced at T. If QR = SR, prove that

PR * ST² = UR * RT²

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Answer:

angle RSQ = angle SQR (isos)
angle SQR = angle SRT (alt. segment)
Hence angle RSQ = angle SQR
==> SQ is parallel to RT
==> PSU and PTR are similar triangles

Using intercept theorem,
UR / PR = ST / PT
UR / PR = ST² / (PT)(ST)

Using tangent secant theorem,
ST * PT = RT²

Hence,
UR / PR = ST² / RT²
PR * ST² = UR * RT²

O lvl A Maths: Plane Geometry

Question from http://www.sgforums.com/forums/2297/topics/359768

In the triangle ABC, BF and CE are perpendicular to AC and AB respectively. D and G are the midpoints of EF and BC respectively. Prove that GD is perpendicular to EF.

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Answer:

dot a perpendicular line from G to BE and label it as H
dot a perpendicular line from G to FC and label it as I

Note that by similar triangles (or by intercept theorem),

Considering triangles HBG and EBC
because BG = ½BC,
EH = ½BE
HG = ½EC


Considering triangles GIC and BFC
because BG = ½BC,
FI = ½FC
GI = ½BF

Thus,
EG² = EH² + HG²
= ¼BE² + ¼EC²
= ¼BC²

GF² = GI² + FI²
= ¼BF² + ¼CF²
= ¼BC²
= EG²

Thus, GF = EG

Since GF = EG, and ED = DF, then GEF is an isosceles triangle where GD is perpendicular to EF