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arpan1

each angle bisector of a triangle divides the opposite side into segments proportional in length to adjacent sides

mukundmadhav

Consider a triangle ABC
Let AD be the angle bisector of angle A
Area of ADB = 1/2 AB. AD sin A/2 = 1/2 BD. AD. sin ADB
Area of ACD= 1/2 AC. AD sin A/2 = 1/2 CD. AD. sin ADC

Now sin ADC = sin ADB
Because ADB = 180 - ADC
Now divide the first equation by the second to get your result

arpan1

from where u got

Area of ADB = 1/2 AB. AD sin A/2 = 1/2 BD. AD. sin ADB

Area of ACD= 1/2 AC. AD sin A/2 = 1/2 CD. AD. sin ADC

that turns out to be 1/2 * base * height which is valid only for perpendicular bisectors

1/2 AB * AD *SIN A/2 = 1/2 AD* BD

mukundmadhav

It is 1/2 side. side . sin of angle between them..

mukundmadhav

1/2absinC

allamraju

Are yaar,He used area=1/2bcsinA=1/2casinB=1/2absinC.He is correct,It is called vertical angle bisector theorem.I remember that I have done it in 10th class.

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Are yaar,He used area=1/2bcsinA=1/2casinB=1/2absinC.He is correct,It is called vertical angle bisector theorem.I remember that I have done it in 10th class.