Area of Triangle ABC:
Let there be a triangle ABC and draw a perpendicular AD on the side BC from A
So, now D = ½ AD.BC.
In triangle ABD
If in a triangle ABC, point D divides BC in the ratio m: n and ÐADC =Q then,
1) (m+n) CotQ = mCotα-nCotβ
2) (m+n) CotQ= nCotβ-mCotC
Find the value of y in
So, y = 0.
The Circle passing through points A, B, C of a triangle ABC, its radius is denoted by R.
Bisect the 2 side BC and AC in D and E respectively and draw DO and EO perpendicular to BC and CA
So, O is center of circumcircle.
Now DBOD @DDOC (By RHS congruency rule)
So, ÐBOD = ÐDOC
= ½ ÐBOC
Now, in DBOD, BD = BO Sin (ÐBOD)