$$ A B C $$ is a triangle right angled at $$ C $$. $$ A $$ line through the mid-point $$ M $$ of hypotenuse $$ A B $$ and parallel to $$ B C $$ intersects $$ A C $$ at $$ D $$. Then prove that, i. $$ D $$ is the midpoint $$ A C $$ ii. $$ M D $$ is perpendicular to $$ A C $$ ii. $$ C M=A M=\frac{1}{2} A B $$

Question

UpStudy AI · Accepted Answer

To prove the statements about triangle $$ ABC $$ with right angle at $$ C $$, we will use coordinate geometry. Let's place the triangle in the coordinate plane for convenience.

1. **Assign Coordinates:**
   - Let $$ C = (0, 0) $$
   - Let $$ A = (a, 0) $$ (on the x-axis)
   - Let $$ B = (0, b) $$ (on the y-axis)

2. **Find the Midpoint $$ M $$ of Hypotenuse $$ AB $$:**
   The coordinates of $$ M $$ can be calculated as:
   $$
   M = \left( \frac{a + 0}{2}, \frac{0 + b}{2} \right) = \left( \frac{a}{2}, \frac{b}{2} \right)
   $$

3. **Equation of Line $$ AC $$:**
   The slope of line $$ AC $$ is:
   $$
   \text{slope of } AC = \frac{0 - 0}{a - 0} = 0
   $$
   Thus, the equation of line $$ AC $$ is simply $$ y = 0 $$.

4. **Equation of Line $$ BC $$:**
   The slope of line $$ BC $$ is:
   $$
   \text{slope of } BC = \frac{b - 0}{0 - 0} = \text{undefined}
   $$
   Thus, the equation of line $$ BC $$ is $$ x = 0 $$.

5. **Line through $$ M $$ parallel to $$ BC $$:**
   Since line $$ BC $$ is vertical, the line through $$ M $$ parallel to $$ BC $$ will also be vertical. The equation of this line is:
   $$
   x = \frac{a}{2}
   $$

6. **Finding Intersection Point $$ D $$:**
   To find point $$ D $$, we need to find the intersection of the line through $$ M $$ (which is vertical) with line $$ AC $$ (which is horizontal). Since line $$ AC $$ is at $$ y = 0 $$, the coordinates of $$ D $$ will be:
   $$
   D = \left( \frac{a}{2}, 0 \right)
   $$

### Proving the Statements

**i. Prove that $$ D $$ is the midpoint of $$ AC $$:**
The coordinates of $$ A $$ are $$ (a, 0) $$ and $$ C $$ are $$ (0, 0) $$. The midpoint $$ P $$ of segment $$ AC $$ is given by:
$$
P = \left( \frac{a + 0}{2}, \frac{0 + 0}{2} \right) = \left( \frac{a}{2}, 0 \right)
$$
Since $$ D = P $$, we conclude that $$ D $$ is the midpoint of $$ AC $$.

**ii. Prove that $$ MD $$ is perpendicular to $$ AC $$:**
The line $$ AC $$ is horizontal (slope = 0), and the line $$ MD $$ is vertical (undefined slope). Since a vertical line is perpendicular to a horizontal line, we conclude that $$ MD $$ is perpendicular to $$ AC $$.

**iii. Prove that $$ CM = AM = \frac{1}{2} AB $$:**
First, we calculate the length of $$ AB $$:
$$
AB = \sqrt{(a - 0)^2 + (0 - b)^2} = \sqrt{a^2 + b^2}
$$
Now, we find $$ CM $$:
$$
CM = \sqrt{\left( \frac{a}{2} - 0 \right)^2 + \left( \frac{b}{2} - 0 \right)^2} = \sqrt{\left( \frac{a^2}{4} + \frac{b^2}{4} \right)} = \frac{1}{2} \sqrt{a^2 + b^2} = \frac{1}{2} AB
$$
Next, we find $$ AM $$:
$$
AM = \sqrt{\left( \frac{a}{2} - a \right)^2 + \left( \frac{b}{2} - 0 \right)^2} = \sqrt{\left( -\frac{a}{2} \right)^2 + \left( \frac{b}{2} \right)^2} = \sqrt{\frac{a^2}{4} + \frac{b^2}{4}} = \frac{1}{2} \sqrt{a^2 + b^2} = \frac{1}{2} AB
$$

Thus, we have shown that $$ CM = AM = \frac{1}{2} AB $$.

### Conclusion
We have proven all three statements:
1. $$ D $$ is the midpoint of $$ AC $$.
2. $$ MD $$ is perpendicular to $$ AC $$.
3. $$ CM = AM = \frac{1}{2} AB $$.
$$ D $$ is the midpoint of $$ AC $$, $$ MD $$ is perpendicular to $$ AC $$, and $$ CM = AM = \frac{1}{2} AB $$.

Solución de tutoría real

Responder

Solución

The Deep Dive

preguntas relacionadas

Latest Geometry Questions