Given: Circle k (O), O ∈ AB
CD altitude to AB
Prove: AD·CB = AC·CD
Why did the circle go to the doctor? Because it had too many degrees! 😄
Ok, let's get serious for a moment. We need to prove that AD·CB = AC·CD. To do that, let's use triangles and their properties.
Since CD is an altitude to AB, it means that CD is perpendicular to AB. So we have a right triangle ACD.
Now let's consider triangle ABC. Since CD is perpendicular to AB, it means that it is also perpendicular to the line segment BC. This makes triangle BCD also a right triangle.
So we have two right triangles, ACD and BCD, sharing the side CD.
Now, let's use a little trick, the Pythagorean theorem! In triangle ACD, we have the following equation:
AC^2 = AD^2 + CD^2
And in triangle BCD, we have:
BC^2 = BD^2 + CD^2
But since we want to prove AD·CB = AC·CD, let's rewrite the equation for AC:
AC^2 = (AD^2 + CD^2) + (BD^2 + CD^2)
AC^2 = AD^2 + BD^2 + 2CD^2
Now let's go back to the problem: AD·CB = AC·CD.
If we rewrite the equation for AD·CB, we get:
AD·CB = AD·(BD + CD)
AD·CB = AD·BD + AD·CD
Now let's substitute in AD^2 + BD^2 = AC^2 - CD^2 (from the Pythagorean theorem) and we get:
AD·CB = (AC^2 - CD^2) + AD·CD
AD·CB = AC^2 + AD·CD - CD^2
And if we compare this equation with the earlier result, we see that they are the same!
Therefore, we have proved that AD·CB = AC·CD. And now, if you have any more geometry questions, I'm your jester-tained go-to bot! 😁
To prove that AD · CB = AC · CD, we will use the property of triangles and the Pythagorean theorem. Here's a step-by-step solution:
Step 1: Draw the given figure with circle k (O), O ∈ AB, and CD as the altitude to AB.
Step 2: Let's consider triangle ABC. Since CD is an altitude, it intersects AB at a right angle. So, we have a right triangle.
Step 3: Using the Pythagorean theorem, we can say that AC^2 = AD^2 + CD^2 and BC^2 = CD^2 + BD^2.
Step 4: Rearranging the equations, we get AD^2 = AC^2 - CD^2 and BD^2 = BC^2 - CD^2.
Step 5: Now, we multiply the first equation by BC^2 and the second equation by AC^2: AD^2 · BC^2 = (AC^2 - CD^2) · BC^2 and AC^2 · BD^2 = AC^2 · (BC^2 - CD^2).
Step 6: Expanding the equations, we get AD^2 · BC^2 = AC^2 · BC^2 - CD^2 · BC^2 and AC^2 · BD^2 = AC^2 · BC^2 - AC^2 · CD^2.
Step 7: Subtracting the second equation from the first, we have AD^2 · BC^2 - AC^2 · BD^2 = CD^2 · BC^2 - AC^2 · CD^2.
Step 8: Factoring out CD^2 from the right side, we get AD^2 · BC^2 - AC^2 · BD^2 = CD^2 · (BC^2 - AC^2).
Step 9: Since we know AD · BC = AC · BD (by the property of triangles), we can rewrite the equation as (AD · BC)^2 - (AC · BD)^2 = CD^2 · (BC^2 - AC^2).
Step 10: We can further simplify the equation as (AD · BC + AC · BD) · (AD · BC - AC · BD) = CD^2 · (BC^2 - AC^2).
Step 11: Dividing both sides by AD · BC + AC · BD, we get AD · BC - AC · BD = CD^2 · (BC^2 - AC^2) / (AD · BC + AC · BD).
Step 12: We need to show that (AD · BC + AC · BD) is not equal to 0. Since AD and BC are sides of the triangle, they cannot be zero. Thus, AD · BC + AC · BD ≠ 0.
Step 13: Dividing both sides by AD · BC - AC · BD, we have AD · CB = CD^2 · (BC^2 - AC^2) / (AD · BC - AC · BD).
Step 14: We need to show that CD^2 · (BC^2 - AC^2) / (AD · BC - AC · BD) is not equal to zero. Since CD, BC, and AC are all non-zero lengths (given in the figure), CD^2 · (BC^2 - AC^2) ≠ 0.
Step 15: Therefore, AD · CB = CD^2 · (BC^2 - AC^2) / (AD · BC - AC · BD) ≠ 0, which proves AD · CB = AC · CD.
This completes the proof.
To prove the given statement AD·CB = AC·CD, we need to use the properties of similar triangles.
1. Draw a diagram: Start by drawing a circle with center O, and draw a line AB that intersects the circle at point O. Let CD be the perpendicular altitude from point C to line AB.
2. Identify similar triangles: We can see that triangle ACD and triangle BCO are similar triangles since angle ACD and angle BCO are both 90 degrees (because CD is the altitude), and angle ADC is congruent to angle BOC as they are both inscribed angles subtending the same arc. This means that the ratio of their corresponding sides must be equal.
3. Write the ratios: Let's denote the lengths of the sides as follows:
AD = a
CB = b
AC = x
CD = y
The ratios of the corresponding sides are:
AC / AD = x / a (from triangle ACD)
CB / CO = b / y (from triangle BCO)
4. Equate the ratios: Since the triangles are similar, we can equate the ratios:
x / a = b / y
5. Cross-multiply: Multiply both sides of the equation by 'a' and 'y' respectively to eliminate the denominators:
x * y = a * b
6. Rearrange the equation: We get:
a * b = x * y
This shows that AD·CB = AC·CD, which proves the given statement.
7. Q.E.D: Therefore, we have successfully proved that AD·CB = AC·CD using the properties of similar triangles.
rewrite the equation as
AD/CD = AC/BC
Since the two small triangles are similar, this must be true.