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Internal and External bisector of an Angle of a Triangle
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Internal and External bisector of an Angle of a Triangle
Theorem
In a triangle ABC, if D is a point on BC such that D divides BC in the ratio AB:AC, then AD is the bisector of the A
              OR

If a line through one vertex of a triangle divides the opposite sides in the ratio of other two sides, then the line bisects the angle at the vertex.
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Example: In a quadrilateral ABCD, if bisectors of the ABC and ADC meet on the diagonal AC, prove that the bisectors of BAD and BCD will meet on the diagonal BD.
GIVEN ABCD is a quadrilateral in which the bisectors of ABC and ADC meet on the diagonal AC at P.
TO PROVE Bisectors of BAD and BCD meet on the diagonal BD
CONSTRUCTION Join BP and DP. Let the bisector of BAD meet BD at Q. Join AQ and CQ.
PROOF  In order to prove that the bisectors of BAD and BCD meet on the diagonal BD. It is sufficient to prove that CQ is the bisector of BCD. For which you will prove that Q divides BD in the ratio BC: DC.
In ABC, BP is the bisector of ABC
 AB/BC = AP/PC......(i)
In ACD, DP is the bisector of ADC
AD/DC = AP/PC......(ii)
From equations (i) and (ii), you get
     AB/BC = AD/DC
AB/AD = BC/DC......(iii)
in ABD ,
AQ is the bisector of BAD [By construction]
AB/AD = BQ/DQ......(iv)
from equation (iii) and (iv) you get
BC/DC = BQ/DQ
thus CBD , Q divides BD in the ratio CB:CD. Therefore, CQ is the bisectors of BCD.
Hence, bisector of BAD and BCD met on the diagonal BD.
 
   
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