Question

(1) In the adjacent figure, chords
AB and CD of a circle with
centre P intersect each other at
point E. Chords AD and CB
are drawn.
P.​

Answer 1

AE × EB = CE × ED (Proved)

Step-by-step explanation:

We have to prove that, AE × EB = CE × ED

Now, in Δ CEB and Δ AED,

∠ CEB = ∠ AED {They are vertically opposite angles}

∠ BCE = ∠ EAD {Those are the angles inscribed in the same arc}

So, Δ CEB ~ Δ AED {A-A test of similarity}

So, $\frac{CE}{AE} = \frac{BE}{ED}$  

{The corresponding sides of two similar triangles will be in the same ratio}

⇒ AE × EB = CE × ED (Proved)

Answer 2

AE × EB = CE × ED (Proved)

Step-by-step explanation:

We have to prove that, AE × EB = CE × ED

Now, in A CEB and A AED,

< CEB = AED (They are vertically opposite angles}

< BCE = 4 EAD (Those are the angles

inscribed in the same arc}

So, A CEBA AED (A-A test of similarity}

CE So, C AE

BE ED

{The corresponding sides of two similar triangles will be in the same ratio}

→ AE EB = CE × ED (Proved)