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Q45.

Expert-verifiedFound in: Page 290

Book edition
Student Edition

Author(s)
Ray C. Jurgensen, Richard G. Brown, John W. Jurgensen

Pages
227 pages

ISBN
9780395977279

**The arithmetic mean between two numbers $r$ and $s$ is defined to be $\frac{r+s}{2}$.**

**$\left(a\right).$ $\overline{CM}$ is the median and $\overline{CH}$ is the altitude to the hypotenuse of right $\Delta ABC$. Show that $CM$ is the arithmetic mean between $AH$ and $BH$, and that $CH$ is the geometric mean between $AH$ and $BH$. Then use the diagram to show that the arithmetic mean is greater than the geometric mean.**

**$\left(b\right).$ Show algebraically that the arithmetic mean between two different numbers $r$ and $s$ is greater than the geometric mean.**

**$\left(a\right).$**

**$\overline{CM}$ is the arithmetic mean between $AH$ and $BH$, and that $CH$ is the geometric mean between $AH$ and $BH$ and arithmetic mean is greater than geometric mean.**

Following diagram is given,

Sum of angles of a triangle are $180\xb0$.

As $CM$ is the median at right angle of $\Delta ABC$, if a perpendicular is drawn from the median at hypotenuse to any other side it will also be the median of triangle formed.

That is $\overline{PM}$ is the altitude as well as median of $\Delta BMC$ at point $P$.

Now consider $\Delta BMP$and $\Delta CMP$,

Here *MP* is the common side, $\angle CPM=\angle BPM=90\xb0$and $\overline{CP}=\overline{BP}$.

Thus, by side-angle-side theorem of congruency,

$\Delta CMP\cong \Delta BMP$

Hence, $\overline{CM}=\overline{BM}$….(i) as they are corresponding sides of triangle considered.

Also, $\angle MCP=\angle MBP=\alpha $.

Now, consider $\Delta ACM$, here $\angle ACM=90-\alpha $

In $\Delta ABC$, sum of all angles in a triangle is $180\xb0$,

therefore

$\begin{array}{c}\angle A+\angle B+\angle C={180}^{\circ}\\ \angle A+{90}^{\circ}+\alpha ={180}^{\circ}\\ \angle A={90}^{\circ}-\alpha \end{array}$

As a result, it can be said that, $\angle A=\angle ACM.$

Now, in $\Delta ACM$, as $\angle A=\angle ACM$, therefore the side opposite to these angle will also be equal.

Hence, $\overline{CM}=\overline{AM}$…(ii)

$\begin{array}{c}2CM=AM+BM\\ 2CM=AB\\ 2CM=AH+BH\\ CM=\frac{AH+BH}{2}\end{array}$

Thus, arithmetic mean of $AH$ and $BH$ can be defined as $CM$

For geometric mean,

Here, consider $\Delta AHC$and $\Delta CHB$,

$\angle AHC=\angle CHB=90\xb0$

Side $AH$ is common.

Also, if $\angle A=x$, then $\angle ACH=90\xb0-x$ and $\angle HCB=x$

Thus, $\angle HCB=\angle ACH$

Hence, by angle-side-angle theorem of similarity, $\Delta AHC\approx \Delta CHB$

Therefore, the ratio of their sides will be,

$\begin{array}{c}\frac{CH}{AH}=\frac{BH}{CH}\\ {\left(CH\right)}^{2}=\left(AH\right)\left(BH\right)\\ \left(CH\right)=\sqrt{\left(AH\right)\left(BH\right)}\end{array}$

Thus, it is proved that $CH$ is the geometric mean of $AH$and $BH$.

Now, to prove that arithmetic mean is greater than the geometric mean using diagram one can simply refer to the diagram as

Here as proved $CM$ is the arithmetic mean and $CH$ is geometric mean.

To prove this by diagram, the given triangle is a right angle triangle with altitude and median of different length thus it is not an isosceles triangle.

The definition of altitude is that it is the smallest distance from an angle to the side opposite to it or it is the height of the triangle from the angle through which the altitude passes.

Whereas median is line joining a vertex of a triangle and the mid-point of the side opposite to it.

As clearly indicated in the figure that altitude and median are different for the given triangle, median is larger than altitude.

That is $CM$ is larger than $CH$*, *which proves that arithmetic mean is larger than geometric mean.

Therefore,

Arithmetic mean is larger than geometric mean can be proved by geometry using basic properties of a triangle and can also be proved by observing the figure given and definition of altitude and median.

$\left(b\right).$

** The arithmetic mean between two different numbers $r$ and s is greater than the geometric mean.**

Two positive numbers, r and s.

Arithmetic mean: $\frac{r+s}{2}$

Geometric mean: $\sqrt{\mathbf{r}\mathbf{s}}$

For two numbers r and s.

${(r-s)}^{2}$ is always positive.

Consider,

${(r-s)}^{2}\ge 0$

${r}^{2}-2rs+{s}^{2}\ge 0$

Since r and s. are positive,

${r}^{2}-2rs+{s}^{2}+4rs\ge 0+4rs$

${r}^{2}+2rs+{s}^{2}\ge 4rs$

${(r+s)}^{2}\ge 4rs$

$r+s\ge 2\sqrt{rs}$

$\frac{r+s}{2}\ge \sqrt{rs}$

Therefore, arithmetic mean is greater than the geometric mean.

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