Animated Solution for Mathematics - Trigonometry: In a $\Delta ABC$, $\angle A=90^{\circ }$ and $AD$ is an altitude. Complete the relation $\frac{BD}{BA}=\frac{AB}{(\dots )}$

Imagine you are standing in a vast, open field, and before you lies a perfect right-angled triangle, $\Delta ABC$. The angle at vertex $A$ is a crisp, precise $90^{\circ }$. This is our foundation, a structure defined by its simplicity and symmetry.

Now, let us introduce an altitude, $AD$, dropping from vertex $A$ straight down to the hypotenuse $BC$. This single line, $AD\perp BC$, transforms our single triangle into a complex, interconnected system of smaller right-angled triangles.

We are tasked with completing the relation:

At first glance, this might look like a simple algebraic puzzle, but it is actually a window into the profound concept of triangle similarity.

When we drop that altitude $AD$, we create two smaller triangles, $\Delta BDA$ and $\Delta ADC$, both nestled within the larger $\Delta ABC$. To solve our ratio, we must look at the relationship between the small triangle $\Delta BDA$ and the original, large triangle $\Delta BAC$.

Look closely at angle $B$. It is common to both the small triangle $\Delta BDA$ and the large triangle $\Delta BAC$. It is the anchor of our logic.

Furthermore, we know that $\angle BDA=90^{\circ }$ because $AD$ is an altitude, and $\angle BAC=90^{\circ }$ because it was given in the problem. With two pairs of equal angles, we invoke the powerful Angle-Angle (AA) Similarity Criterion.

This tells us that $\Delta BDA\sim \Delta BAC$. The triangles are not identical in size, but they are identical in shape, meaning their sides are proportional.

Now that we have established $\Delta BDA\sim \Delta BAC$, we can write the ratios of their corresponding sides. We must match the vertices carefully:

We do not need the middle term involving $DA$ and $AC$. We only need the first and the last:

Since $BA$ is the same as $AB$, we can rewrite this as:

Comparing this to our original target, $\frac{BD}{BA}=\frac{AB}{(\dots )}$, the answer reveals itself: the missing denominator is $BC$.

This is not just an answer; it is the Geometric Mean Theorem in action, which states that:

You have just navigated the heart of geometric similarity. Keep this logic in your toolkit; it is a powerful way to see the hidden relationships in any triangle.