Question

Please help me with this...:)



Please

Answer 1

see the attached picture

Answer 2

Hi there!

✨ Given: AB and CD are two chords of the circle with center O, which intersects at E.

✨ To prove:  ∠AEC = 1/2 (∠COA+∠DOB)

✨ Construction:  join AC, AO , OC , BC and BD

✨ Proof:
Since, Angle subtended at center is double the angle subtended at circumference.

AC is Chord:
∠AOC = 2∠ABC ----(i)

Similarly, BD is Chord
∠DOB = 2∠DCB ----(ii)

Adding Eqn. (i) & (ii)

∠AOC + ∠DOB = 2(∠ABC + ∠DCB) ----(iii)

In triangle CEB,

∠AEC = ∠ECB +∠CBE [Exterior angle is the sum of two opposite interior angles]

∠AEC = ∠DCB + ∠ABC ----(iv)

From Eqn. (iii) & (iv)

∠AOC +∠DOB = 2∠AEC
Or,

∠AEC = 1/2 × (∠AOC +∠DOB)

[HENCE PROVED]

Cheers!