"Strong is this woman's mighty 阴阳眼, from the quick working-less way she obtained her prove merely "with some simple substitution and factorisation ... hee hee""

To prove the seemingly elusive PE² = PD×PC − DE×CE we first list out the relevant properties/theorems that will contain the our required line segments, as JOYL has done:

And now this is the moment when one hopes that the 阴阳眼 will come. And often it comes with the most simplest and basic of properties that you've learnt in lower secondary!

Like in this, students are so caught up trying to recall the latest theorems that they often forget about ME - the Pythagoras Theorem! Add my fellow old friends like interior angles of triangles and parallel lines to the list as well!

So ...

Now it's time for JOYL's "simple substitution and factorisation ... hee hee" part (please correct Miss Loi if this isn't your exact method ok?) ...

Now look closely at the chord AB in the diagram and using equation (2):
DE × CE
=BE × EA
= (BM+ME)(MA−ME)
= (MA+ME)(MA−ME) since BM=MA as M is mid-pt. of AB (given)
= MA²−ME²

So sub DE × CE = MA²−ME² into (3),
PA² − PE² = DE × CE²
PE² = PA² − DE × CE²

And sub PA² = PD × PC from (1):
PE² = PD × PC − DE × CE (Yay!)

So you see, one of the most crucial part of the working is being able to express PE using the Pythagoras Theorem, and recognizing that I'm not some old forgotten has-been who still is useful in Plane Geometry society! So please don't retrench me!

"Now go my Sexy Maths Tutor!" says the Right-Angled Triangle as he handed her the latest copy of the Plane Geometry Street Directory.

"You follow this straight line ... then turn left at the intersection of the two chords ... then follow this tangent ... "

Here's an alternative proof just for the sake of it:

• ∠BMO = ∠PMA (vert. opp. ∠s)

• Let ∠MPA = θ
  ∠PMA = 90° (⊥ bisector of Chord AB)

  Take ΔPMA,
  ∠PAM = 180° − 90° − θ = 90° − θ

• PA = PB (Tangents from ext. pt.)
  ⇒ ΔPBA is isosceles
  ⇒ ∠PBA = ∠PAB = 90°−θ (base ∠s of isosceles Δ)

• ∠PBO = 90° (Radius ⊥ Tangent)
  ⇒ ∠MBO = ∠PBO − ∠PBM = 90°−90°−θ = θ
  ⇒ ∠APM = ∠MBO

∴ ΔPMA is similar to ΔBMO (2 angles similar)

Now where's Part 3???! Miss Loi's idea of a long weekend doesn't including spending it in front of a grumpy right-angled triangle!

And since the two triangles are similar,

cross multiplying, PM x BO= PA x BM (shown)

Miss Loi: Nothing like a diagram to brighten up your workings and a National Day

• Angle PMA= Angle BMO (vert. opp. angles)

• let these 2 angles be y.
  Therefore, Angle BMP= Angle AMP= 180-y

  Think you'd agree this part is unnecessary for the final working.

• Angle PBO= 90 degrees (tan. perpen. rad)

• let angle ABO= x, hence Angle BAO= x
  (BO=AO cos they are both radii, hence triangle ABO is isosceles and base angles are equal.) Angle BOA= 180-2x

  (Ok Miss Loi, this is the part where I am not sure. Please point out if there's flawed reasoning... D:)

  Alright may this little diagram on the right (taken from the official Jφss Sticks Plane Geometry cheatsheet) help firm up your reasoning to a steely level.

  

  Given that M is the mid-point of AB, and recalling the Perpendicular Bisector of a Chord property which most of you would've learnt in your Geometric Properties of Circles, the line joining the circle center O to M will imply that ∠BMO = 90°.

  And by taking the interior angles of ΔBMO we can obtain ∠BOM straightaway as 180° − 90° − x = 90° − x = ∠PAM.

  P.S. This in turn will also confirm chop stamp prove that PO bisects ∠BAO, since PAOB is actually a layang-layang kite.

  Angle PAO= 90 degress (tan. perpen. rad)
  And since Angle BAO= x, Angle PAM= 90-x.

  PO bisects Angle BAO, hence angle BOM= (180-2x)/2
  = 90-x.
  Therefore Angles PAM and BOM are the same and both triangles PMA and BMO are similar (AA)

Hence, shown. Yay! One step closer to Miss Loi's liberation ...