# solution-of-triangles-3

** Area** **of Triangle ABC: **

How?

Fig (6)

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

Why?

Solution:

__m-n theorem __

__ __

If in a triangle ABC, point D divides BC in the ratio m: n and ÐADC =Q then,

Fig (7)

1) (m+n) CotQ = mCotα-nCotβ

2) (m+n) CotQ= nCotβ-mCotC

**Illustration 5: **

Find the value of y in

?

Ans:

So, y = 0.

**CIRCUM****CIRCLE: **

The Circle passing through points A, B, C of a triangle ABC, its radius is denoted by R.

How?

Bisect the 2 side BC and AC in D and E respectively and draw DO and EO perpendicular to BC and CA

Fig (8)

So, O is center of circumcircle.

Now DBOD @DDOC (By RHS congruency rule)

So, ÐBOD = ÐDOC

= ½ ÐBOC

= ÐBAC

= ÐA

Now, in DBOD, BD = BO Sin (ÐBOD)