# [solved] dissecting an isosceles triangle

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1. May 7, 2015

### stlukits

Simple question, but I can't figure it out. Consider an isosceles triangle ABC with \alpha=\beta dissected by a line through C and D, where D is on AB. It is obvious that |CD|<=|AC|=|BC|, but I want to prove it using trigonometry. I can use |BD|<=|BC| in my assumptions but not angle(BCD)<=angle(ACB). Otherwise I'd be done, for the law of sines using the triangle ABC and BCD gives me |BC|>=|CD|*(sin(angle(BCD))/sin(angle(ACB))). But how to show that angle(BCD)<=angle(ACB), obviously without using what we are trying to prove, i.e. |CD|<=|BC|.

2. May 7, 2015

### HallsofIvy

Staff Emeritus
The bisector, CD, of AB divides the isosceles triangle into two right triangles with AC and BC as hypotenuses, AD and BD as legs. By the Pythagorean theorem, the hypotenuse of a right triangle is always larger than either leg.

3. May 7, 2015

### stlukits

I am not BISECTING but DISSECTING, so congruence between the dissecting triangles is not necessary. Upon further thought, though, angle(ACB)=angle(BCD)+angle(DCA) and since angle(DCA)>=0, we get angle(ACB)>=angle(BCD) and thus, by above-mentioned law of sines identity, |CD|<=|BC|.