# solution-of-triangles-4

**Illustration 6:**** **

If in the DABC, O is circumcenter and R is the circumradius and R_{1}, R_{2}, R_{3}, are circumradii of the triangle OBC, OCA, OAB respectively then prove that

Ans:

Fig (9)

Clearly in D OBC, ÐBOC = 2A,

__ __

__Incircle: __

The Circle which touches all the sides of DABC internally its radius is denoted by r.

How?

Bisect the ÐB and ÐC by line BI and CI meeting in I

So now I is in center of the circle. Draw ID, IE and IF ^ to 3 sides.

So, ID = IE = IF = r

Fig (10)

Now we have

Area of DIBC = ½ ID.BC= ½ r.a

Area of DIAC = ½ IE.AC= ½ r.b

Area of DIAB = ½ IF.AB= ½ r.c

Area of DABC= area of DIBC + area of DIAC + area of DIAB.

Now what about r =

__Illustration____7: __

A, B, C are the angles of a triangle, prove that:

Ans:

__Excircles or Escribed circles: __

The circle which touch BC and the two sides AB and AC produced is called escribed circle. Opposite the angle A and its radius is denoted by r_{1}. Similarly radii of circle opposite the angle B and C are denoted by r_{2} and r_{3} respectively.

How?

Fig (11)

Produce AB and AC to L and M. Bisect ÐCBL and ÐBCM by lines BI and CI and, let these lines meet in I.

Draw ID, IE, IF ^ to 3 sides respectively.

I is center of the escribed circle and so ID=IF=IE=r_{1}.

Now area of DABC + area of DIBC = area of DIAB + area of DIAC

So, D+ ½ (ID) (BC) = ½ (IF) (AB) + ½ (IE) (AC)

Þ D + ½ r_{1}a = ½ r_{1}c + ½ r_{1}b

So, D = ½ r_{1} (b+c-a)

= ½ r_{1} ((a+b+c)-2a)

= ½ r_{1} (2s-2a)

= r_{1} (s-a)

Now what about

Illustration 8:

Show that

__ __

__Orthocenter and Pedal ____D____: __

Let ABC be any triangle and let AD, BE and CF be the altitudes of DABC. Then the DDEF formed by joining point D, E and F the feet of ^ is called the pedal D of DABC.

Fig (12)

Now

1) AH = 2RCosA, BH = 2RCosB, CH = 2CosC

How?

2) HD = 2R CosB CosC, HE = 2R CosA CosC, HF = 2R CosA CosB

Why?

Fig (14)

HD = BD tan ÐHBD

= BD tan (90^{0 }- c)

= AB CosB.CotC