Question

Given: Chords \( AB \) and \( CD \) of a circle with center \( P \) intersect at point \( E \).
To Prove: \( AE \times EB = CE \times ED \). 

Hint:

For intersecting chords in a circle, use the inscribed angle theorem and properties of similar triangles to establish proportionality and cross-multiplication.

Answer 1

Draw segment \( AC \) and segment \( BD \).
Proof: Step 1: In \( \triangle CAE \) and \( \triangle BDE \), \[ \angle AEC = \angle DEB \quad \text{(angles inscribed in the same arc)}. \] Step 2: Similarly, \[ \angle CAE = \angle BDE \quad \text{(angles inscribed in the same arc)}. \] Step 3: By AA similarity criterion, \( \triangle CAE \sim \triangle BDE \).
Step 4: From the property of similar triangles, we have: \[ \frac{AE}{CE} = \frac{EB}{ED}. \] Step 5: Cross-multiplying, we get: \[ AE \times EB = CE \times ED. \] Hence proved.