# Relation between the GM of two sides of a triangle and the bisector of angle between them

I am trying to solve the following multiple choice problem :

Let the bisector of the angle $C$ of a triangle $ABC$ intersect the side $AB$ at a point $D$. Then the geometric mean of $CA$ and $CB$

1. is less than $CD$
2. is equal to $CD$
3. is greater than $CD$
4. does not always satisfy any one of the foregoing properties.

I have tried that following steps :

Since $CD$ bisects $\angle C$, so $\angle ACD=\angle DCB$ and $\dfrac{CA}{CB}=\dfrac{AD}{DB}$.

Also, $ext\angle ADC=\angle DCB+\angle DBC$, so that $\angle ADC>\angle DCB=\angle ACD$. This shows that $CA>AD$. Similarly, $CB>DB$. Now, by triangle inequality, $CA+AD>CD$ and $CB+DB>CD$. Thus it follows that $2CA>CD$ and $2CB>CD$ and this gives me $\sqrt{CA\cdot CB}>\dfrac{CD}{2}$. But the answer as given in my textbook is $\sqrt{CA\cdot CB}>CD$ i.e. the option $3$. How do I proceed to solve this problem?

• If $CA=CB$, then their GM is themselves, and it is clear that neither (1) nor (2) can be the answer. – alex.jordan Jun 30 '14 at 7:54
• ya thats true.. – Debashish Jun 30 '14 at 7:56
• A thought that may help. With a circle, when two chords cross, the products of the corresponding subchords are equal to each other. With this GM in mind, it may help to rearrange things so that AC and CB are in line, by flipping one of the triangles. If say you flipped triangle BDC to B'D'C, then make line segment ACB' be one of the chords, with C as the intersection with another chord. If the circle circumscribes ADB', then you just need to show D' is interior to this circle. (Try to draw what I am saying if this is hard to follow.) However showing D' is inside the circle may be just as hard. – alex.jordan Jun 30 '14 at 8:38
• @alex.jordan .. nice idea. Let me try ! – Debashish Jun 30 '14 at 8:40
• @alex.jordan .. I think I have got a solution. I am typing that out in answer. Thanks for your hint. – Debashish Jun 30 '14 at 8:59

Lemma. $CD<\sqrt{CA\cdot CB}$

Proof.
$\triangle ADC\sim\triangle ABE$ $\Longrightarrow$ $BE=\dfrac{CB+CA}{CA}\cdot CD<CB+CE=2CB$

$\therefore$ $CD<\dfrac{2CA\cdot CB}{CA+CB}\leq\dfrac{2CA\cdot CB}{2\sqrt{CA\cdot CB}}=\sqrt{CA\cdot CB}$

• thanks a lot ! Your solution is so good. $BE=\dfrac{CB+CA}{CA}\cdot CD<CB+CE$ is actually the crucial step ! – Debashish Jun 30 '14 at 10:49

Thanks to alex.jordan for the valuable comment ! I think I have an answer now:

As suggested, I have made the necessary construction. I have produced $AC$ to $AB'$ such that $CB'=CB$ and then produced $DC$ to $DD'$ such that $DC=CD'$. Join $B'$ and $D'$. Then clearly $\Delta BCD\cong \Delta B'CD'$, so that $\angle D'B'C=\angle DBC$. Then I have drawn the circumcircle of $\Delta ADB'$. Let the line joining $D$ and $D'$ intersect this circle at $D_1$. I need to show that $D'$ lies between $D$ and $D_1$. To the contrary let us suppose that $D'$ lies at the position $D_2$ which is outside the circle. Then by construction, $\Delta B'CD_2\cong \Delta BCD$, so that $\angle D_2B'C=\angle D_2B'A=\angle DBC$. But then $\angle D_2B'A>D_1B'A=\angle D_1DA$ which will imply that $\angle DBC>\angle ADC$ but this is not true because $\angle ADC=\angle BDC+\angle DCB$ so that $\angle ADC>\angle DBC$. Thus, $D'$ cannot be outside the circle. similarly we can show that $D'$ cannot lie on the circle too. Hence $D'$ should lie inside the circle. Thus, $CD'<CD_1$. Hence, $CA\cdot CB=CD\cdot CD_1>CD.CD'=CD^2$. Thus, geometric mean of $CA$ and $CB$ is always greater than $CD$.

But, is there any other easier way to solve this problem?