Question

AC and BD are the diagonals of a cyclic quadrilateral ABCD. If the bisector of ✓ACB meets the circle at E, then price that:
(1) DE is bisector of ✓ADB.
(2) ✓BCE=✓ADE

Answer 1

(1) DE is bisector of ∠ADB.

(2) ∠ BCE = ∠ ADE (Proved)

Step-by-step explanation:

See the attached diagram.

Here, CE is the angle bisector of ∠ ACB

Now, join the points E and D.

(1) Now, ∠ ACB = ∠ ADB

{As they are the angles on the circle from equal chord AB}

Now, given that ∠ ACE = ∠ BCE

Again, ∠ BCE = ∠ BDE  ............ (1)

{As they are the angles on the circle from equal chord BE}

Therefore, ∠ BDE = ∠ BCE = 1/2 ∠ ACB = 1/2 ∠ ADB

Therefore, DE is the bisector of ∠ ADB and hence, ∠ ADE = ∠ BDE ............ (2)

(2) From equations (1) and (2) it is proved that ∠ BCE = ∠ ADE.