Question

if two intersecting chords of a circle make equal angles with the diameter passing through their point of intersection, prove that the chords are equal.

Answer 1

Answer:

Step-by-step explanation:

Given that AB and CD are two chords of a circle, with centre O intersecting at a point E.

XY is a diameter passing through E, such that ∠ AEY = ∠ DEY

Construction = Draw OP⊥ AB and OQ ⊥ CD.

Proof-

In right angle DOPE

=∠POE + 90° + ∠ PEO =180° (Angle sum property of a triangle)

=∠POE = 90° – ∠PEO

= 90° – ∠AEY = 90° – ∠DEY = 90° – ∠QEO

= ∠QOE In triangles OPE and OQE, ∠PEO = ∠QEO ∠POE = ∠QOE (Proved) OE = OE (Common side) ∴ ΔOPE ≅ ΔOQE ⇒ OP = OQ (CPCT) Thus, AB = CD

Answer 2

Answer:

Step-by-step explanation:

Given that AB and CD are two chords of a circle, with centre O intersecting at a point E.

XY is a diameter passing through E, such that ∠ AEY = ∠ DEY

Construction = Draw OP⊥ AB and OQ ⊥ CD.

Proof-

In right angle DOPE

=∠POE + 90° + ∠ PEO =180° (Angle sum property of a triangle)

=∠POE = 90° – ∠PEO

= 90° – ∠AEY = 90° – ∠DEY = 90° – ∠QEO

= ∠QOE In triangles OPE and OQE, ∠PEO = ∠QEO ∠POE = ∠QOE (Proved) OE = OE (Common side) ∴ ΔOPE ≅ ΔOQE ⇒ OP = OQ (CPCT) Thus, AB = CD