Q: 4 Let the vertex of an angle ABC be located outside a circle and let the sides of the angle intersect equal chords AD and CE with the circle. Prove that is equal to half the difference of the angles subtended by the chords AC and DE at the centre.
Given : AD = CE
To prove :
Construction: Join AC and DE.
Proof :
Let ADC = x , DOE = y and AOD = z
So, EOC = z (each chord subtends equal angle at centre)
AOC + DOE +AOD + EOC =
.........................................1
In OAD ,
OA = OD (Radii of the circle)
OAD = ODA (angles opposite to equal sides )
OAD + ODA + AOD =
.............................................................2
Similarly,
.............................................................3
..............................................................4
ODB is exterior of triangle OAD . So,
ODB = OAD + ODA
(from 2)
.................................................................5
similarly,
OBE is exterior of triangle OCE . So,
OBE = OCE + OEC
(from 3)
.................................................................6
From 4,5,6 ;we get
BDE = BED = OEB - OED
..................................................7
In BDE ,
DBE + BDE + BED =
...................................................8
Here, from equation 1,
...................................9
From 8 and 9,we have