saketsorcerer
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ABC is a right triangle right angled at B. BD is perpendicular bisector of AC. Find ratio of ar(DBC) and ar(abc). Relate this answer to triangles chapter

Answer :

kvnmurty
Let us take the two triangles  ΔBDA and ΔBDC. These two are congruent as:

DA = DC (as BD is bisecting AC), DB is common, and the
                 Angle BDA = angle BDC = 90  (BD perpendicular to AC)

So It is easy now that:  angle DBA = angle DBC

Since their sum is the angle ABC = 90 deg (given),

   Hence      angle DBC  = 1/2 angle ABC. = 45 deg.

Ratio = 1/2

Mathexpert
Area of triangle ABC = [tex] \frac{1}{2}* AC * BD [/tex]

Area of triangle BDC =  [tex] \frac{1}{2}* CD * BD [/tex]. 

Area of triangle BDC =  [tex]\frac{1}{2}* \frac{1}{2} AC * BD[/tex]

Area of triangle BDC =  [tex]\frac{1}{2}* (\frac{1}{2} AC * BD)[/tex]

[tex]Area of triangle BDC = \frac{1}{2}* (Area of \Delta ABC)[/tex]

So,

[tex] \frac{Ar( \Delta BDC)}{Ar(\Delta ABC)} = \frac{1}{2} [/tex]

The ratio of areas of triangles BDC and ABC is equal to 1 : 2



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