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The Angle Bisector and Equal Side Ratios, Now by Similarity

Date: 01/15/2016 at 10:55:39
From: 
Subject: Relating geometry 

In triangle ABC, AD is the bisector of angle BAC.

Prove geometrically that AB/AC = BD/DC.

This can be proved by trigonometry. But using only geometry, this 
seems hard. 

If the triangles were similar, then there would be a chance of proceeding. 
Here is where I feel the difficulty.



Date: 01/15/2016 at 11:33:32
From: Doctor Floor
Subject: Re: Relating geometry 

Hello Sandip,

Thank you for writing to Dr. Math.

Here is one solution from the Dr. Math archives:

  The Angle Bisector and Equal Side Ratios
    http://mathforum.org/library/drmath/view/55014.html 

But there is another approach as well, using the kind of similar triangles 
that you alluded to.

As a start, extend the angle bisector AD and intersect with a line through 
B parallel to AC. The point of intersection we call E.

Here is an image for your reference:
  

Now note that

- triangles ADC and EDB are similar
- triangle ABE is isosceles

This should help you complete a geometric proof by similarity.

If you need more help, just write back.

Best regards,

- Doctor Floor, The Math Forum
  http://mathforum.org/dr.math/ 
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