Geometry/Angles

ramjmi

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Aug 11, 2012
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plz solve my question it is given that angle B= 2 times angle C and also AB=DC we have to prove that angle BAC=72° and AD is the bisector of angle A.



tri.jpg
 
plz solve my question it is given that angle B= 2 times angle C and also AB=DC we have to prove that angle BAC=72° and AD is the bisector of angle A.



View attachment 2172

To start:

mBAC = 180° - 3*(mBCA)

Please share your work with us indicating exactly where you are stuck - so that we may know where to begin to help you.
 
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To start:

mBAC = 180° - 3*(mBCA)

Please share your work with us indicating exactly where you are stuck - so that we may know where to begin to help you.

Let angle BAC = x.
Than angle ABC = 2x.
and angle BAC = 180 - 3x.

But, Now how to find the value x ??
 
Now let us look at those triangles.

How does the vertex-angle bisector intersect the base - meaning if the line AD is the bisector intersecting base BC - what would be the ratio of BD and DC? If the ratio turns out to be that - does it prove that line is vertex-bisector?

Have you thought through this line?
 
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