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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