triangle, interior angle bisector, circumcenter, incenter

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High School Math Grades 9-12, Ages 13-18

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Valentin Vornicu

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Valentin Vornicu

#1 h Mar 5, 2005, 10:20 PM • 2 Y

Y by Adventure10, Mango247

In the triangle $ABC$ let $AD$ be the interior angle bisector of $\angle ACB$ , where $D\in AB$ . The circumcenter of the triangle $ABC$ coincides with the incenter of the triangle $BCD$ . Prove that $AC^2 = AD\cdot AB$ .

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yetti

#2 h Mar 9, 2005, 5:39 AM • 2 Y

Y by Adventure10, Mango247

Cannnot leave such a nice problem unanswered. Label the sides of the triangle $\triangle ABC$ as $a = BC, b = CA, c = AB$ and its angles at the vertices $A, B, C$ as $\alpha, \beta, \gamma$ . Let $O$ be the common circumcenter of the triangle $\triangle ABC$ and the incenter of the triangle $\triangle BCD$ . Since $O$ is the circumcenter of the triangle $\triangle ABC$ , it lies on the perpendicular bisector of the side $BC$ . Since the incenter of the triangle $\triangle BCD$ lies on the perpendicular bisector of its side $BC$ , this triangle is isosceles with $DB = DC$ , Hence, the angles $\frac \gamma 2 = \measuredangle BCD = \measuredangle CBD = \beta$ are equal. The external angle angle at the opposite vertex $D$ of this isosceles triangle is equal to the sum of these internal angles, that is, $\measuredangle CDA = \frac \gamma 2 + \beta = \frac \gamma 2 + \frac \gamma 2 = \gamma$ . As the sum of the internal angles of the triangle $\triangle ADC$ is $180^o$ ,

$\alpha + \gamma + \frac \gamma 2 = \alpha + \frac 3 2 \gamma = 180^o$

The triangle $\triangle BOD$ is isosceles, because $OB = OC = R$ are radii of the circle $(O)$ . Since $O$ lies on the bisectors of the angles $\measuredangle BCD = \frac \gamma 2$ and $\measuredangle CBD = \beta = \frac \gamma 2$ , the internal angle $\measuredangle BOC$ of this isosceles triangle is equal to

$\measuredangle BOC = 180^o - \left(\frac \beta 2 + \frac \gamma 4 \right) = 180^o - \frac \gamma 2$

But the angle $\measuredangle BOC$ is the central angle of the arc $CAB$ , which means that it is equal to double the angle $\measuredangle CAB = \alpha$ spanning this arc:

$\measuredangle BOC = 180^o - \frac \gamma 2 = 2 \alpha$

Hence, we obtained 2 equations for the angles $\alpha, \gamma$ and these angles can be calculated:

$\alpha + \frac 3 2 \gamma = 180^o$

$2 \alpha + \frac \gamma 2 = 180^o$

$\alpha = \gamma = \frac {360^o}{5} = 72^o$

This means that the triangles $\triangle ABC$ and $\triangle ADC$ are both isosceles with $c = AB = BC = a$ and $CA = CD$ . Since the angle bisector $CD$ of the angle $\gamma = \measuredangle BCA$ cuts the opposite side $c = AB$ in the ratio of the adjacent sides $a = BC, b = CA$ , we have

$\frac{AD}{BD} = \frac b a$

and since $BD = CD = CA = b$

$AD = \frac{b^2}{a}$

Consequently,

$AD \cdot AB = \frac{b^2}{a} \cdot c = \frac{b^2}{a} \cdot a = b^2 = AC^2$

In addition,

$\frac a b = \frac c b = \frac{AB}{CA} = \frac{AB}{BD} = \frac{AD + BD}{BD} = 1 + \frac b a$

and we obtained a quadratic equation for the ratio $\frac a b$ :

$\left( \frac a b \right)^2 - \frac a b - 1 = 0$

$\frac a b = \frac{1 \pm \sqrt 5}{2}$

Since this ratio is positive, the negative root is not acceptable and $\frac a b = \frac{1 + \sqrt 5}{2}$ is equal to the golden ratio.

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mna851

#3 h Mar 9, 2005, 10:33 PM • 2 Y

Y by Adventure10, Mango247

Quote:

Since the incenter of the triangle lies on the perpendicular bisector of its side , this triangle is isosceles with

Is this a well known theorem?

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mna851

#4 h Mar 10, 2005, 2:47 AM • 2 Y

Y by Adventure10, Mango247

Aren't there ways of drawing the perpendicular bisector and angle bisector without having them coincide.

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yetti

#5 h Mar 10, 2005, 5:55 AM • 2 Y

Y by Adventure10, Mango247

mna851 wrote:

Quote:

Since the incenter of the triangle lies on the perpendicular bisector of its side , this triangle is isosceles with

Is this a well known theorem?

This is a simple deduction, too simple to be a theorem. The problem condition states that the point $O$ is the circumcenter of the triangle $\triangle ABC$ . Therefore, the point $O$ lies on the perpendicular bisector of its side $AB$ and the triangle $\triangle BCO$ is isosceles with $OB = OC = R$ , the circumradius. Hence, the angles $\measuredangle BCO = \measuredangle CBO$ are equal. The problem condition also states that the point $O$ is the incenter of the triangle $\triangle BCD$ . This means that the lines $BO, CO$ are the angle bisectors of the angles $\measuredangle CBD, \measuredangle BCD$ . As a result, $\measuredangle CBD = 2 \measuredangle CBO = 2\measuredangle BCO = \measuredangle BCD$ , which means that the triangle $\triangle BCD$ is isosceles.

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kueh

#6 h Mar 10, 2005, 9:27 AM • 2 Y

Y by Adventure10, Mango247

Valentin Vornicu wrote:

In the triangle $ABC$ let $AD$ be the interior angle bisector of $\angle ACB$ ...

Isn't it CD and not AD?

yetti wrote:

mna851 wrote:

Quote:

Since the incenter of the triangle lies on the perpendicular bisector of its side , this triangle is isosceles with

Is this a well known theorem?

This is a simple deduction, too simple to be a theorem. The problem condition states that the point $O$ is the circumcenter of the triangle $\triangle ABC$ . Therefore, the point $O$ lies on the perpendicular bisector of its side $AB$ and the triangle $\triangle BCO$ is isosceles with $OB = OC = R$ , the circumradius. Hence, the angles $\measuredangle BCO = \measuredangle CBO$ are equal. The problem condition also states that the point $O$ is the incenter of the triangle $\triangle BCD$ . This means that the lines $BO, CO$ are the angle bisectors of the angles $\measuredangle CBD, \measuredangle BCD$ . As a result, $\measuredangle CBD = 2 \measuredangle CBO = 2\measuredangle BCO = \measuredangle BCD$ , which means that the triangle $\triangle BCD$ is isosceles.

Once you have come this far, $\measuredangle CBD = \measuredangle BCD = \measuredangle ACD$ Thus by alt seg, $AC$ is tangent to the circumcircle of $CDB$ , so we are done.

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yetti

#7 h Mar 10, 2005, 7:47 PM • 2 Y

Y by Adventure10, Mango247

kueh wrote:

Once you have come this far, $\measuredangle CBD = \measuredangle BCD = \measuredangle ACD$ Thus by alt seg, $AC$ is tangent to the circumcircle of $CDB$ , so we are done.

Very nice, I missed that.

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