Question

in a circle O is the centre ,AB is a diameter and CD is chord equal to the circle .AC and BD when produced intersects at E .Prove that angle AEB =60 degree​

Answer 1

$\green{ \mathfrak{ \large{Given}}}$

• In a circle O is the centre ,AB is a diameter and CD is chord equal to the circle .
• AC and BD when produced intersects at E

$\mathfrak{ \underline{ \underline{We \: have \: to \: prove \: that, }}}$

• ∠CEB=60°

$\purple{ \mathfrak{ \large{ \underbrace{Solution}}}}$

CD= radius r

OC=OD= radius

OCD is equilateral triangle

∠DCO=∠COD=∠ODC=60°

∠ACB=90°(Angle in semicircle)

∠DOC=2∠DBC (half angle)

∠DBC=30°

∠ECB+∠BCA=180° (linear pair)

∠ECB=180−90=90°

In △ECB

∠CEB+∠ECB+∠CBE=180°

∠CEB+90+30=180°

∠CEB=60°

(Proved) ✅

❣Hope this helps you✌