A math - plane geometry help

??aaa??Wwqa SG

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 X AF = AQ X AB

(d) AE = AP

link to diagram:

http://s628.photobucket.com/albums/uu10/finalpig/?action=view&current=untitled.jpg

Forbiddensinner SG

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 X AF = AQ X AB

(d) AE = AP

link to diagram:

http://s628.photobucket.com/albums/uu10/finalpig/?action=view&current=untitled.jpg

Here is your image. Just posting it up so that it will be easier to see.

Forbiddensinner SG

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 X AF = AQ X AB

(d) AE = AP

link to diagram:

http://s628.photobucket.com/albums/uu10/finalpig/?action=view&current=untitled.jpg

I am sorry but I could only figure out the solutions for the first 2 of your questions. I believe the mods will be of a bigger help when they log on. In the mean time, I will continue to study the figure till I can come up with something.

(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

??aaa??Wwqa SG

i know first 3 part but don't know how to do the last part

Diruga Vejida SG

i know first 3 part but don't know how to do the last part

How about you post the solution for the 3rd question for others? Maybe it will be easier for them to solve the 4th question this way.

??aaa??Wwqa SG

AE^2 = AF X AD( tangent-secant theorem)

From part (b) AD X AF = AQ X AB

AE^2 = AQ X AB

AE^2 = AQ ( AQ + AB)

AE^2 = AQ^2 + AQ X AB (proven)

MasterMoogle SG

if can prove pq-square = aq X qb then very easy to do liao

MasterMoogle SG

just a guess, pqa and bqp r similar triangles, so pq/aq = qb/pq and pq-square = aqXqb

ap-square = pq-square+aq-suare = aq-square+aqXqb = ae-square

so ap = ae

shifu or the others lai verify

Forbiddensinner SG

just a guess, pqa and bqp r similar triangles, so pq/aq = qb/pq and pq-square = aqXqb

ap-square = pq-square+aq-suare = aq-square+aqXqb = ae-square

so ap = ae

shifu or the others lai verify

Interesting. So how did you prove that PQA and BQP are similar triangles?

Forbiddensinner SG

I have pmed lecturor Wee, and this is the hint which he has dropped me :

Draw a line to join P and E.

Look at triangle APE from which we wish to show that AP = AE, i.e. APE is isoceles.

Can we show that line AF is perpendicular to PE?

If so, then triangle APE is isoceles.

I still haven't figure out how to prove that AF is perpendicular to PE, though I will continue to think about it.

??aaa??Wwqa SG

never mind i know the answer already

MasterMoogle SG

never mind i know the answer already

cn share pls?

??aaa??Wwqa SG

Draw a full circle and extend a line to form a chord

AD = AQ X QD

PQ = QZ ( radius from centre bisect chord)

PQ X QZ = PQ^2

Therefore: PQ^2 = AQ X QB ( Intersecting chords theorem)

From part (c) : AQ^2 = AE^2 - (AQ)(QB)

AP^2 = AQ^2 + PQ^2

Therefore: AP^2 = [AE^2 - (AQ)(QB)] + [(AQ)(QB)] ( Replaced found eqn)

AP^2 = AE^2

square root them: AP = AE ( PROVEN)

MasterMoogle SG

Draw a full circle and extend a line to form a chord

AD = AQ X QD

PQ = QZ ( radius from centre bisect chord)

PQ X QZ = PQ^2

Therefore: PQ^2 = AQ X QB ( Intersecting chords theorem)

From part (c) : AQ^2 = AE^2 - (AQ)(QB)

AP^2 = AQ^2 + PQ^2

Therefore: AP^2 = [AE^2 - (AQ)(QB)] + [(AQ)(QB)] ( Replaced found eqn)

AP^2 = AE^2

square root them: AP = AE ( PROVEN)

Thanks man!