Animated Solution for Mathematics - Trigonometry: In $\Delta ABC$, internal angle bisector of $\angle A$ meets side $BC$ in $D$. $DE\perp AD$ meets $AC$ in $E$ and $AB$ in $F$. Then

Unlocking the Triangle

Imagine you are standing before a triangle $\Delta ABC$. It looks simple, but hidden within its structure are profound symmetries waiting to be uncovered.

Today, we are going to explore the internal angle bisector $AD$ and a line $EF$ drawn perpendicular to it. This isn't just a geometry problem; it is a masterclass in how we can use symmetry and area to simplify complex relationships.

The Power of Congruence

Let's start by visualizing our construction. We have $\Delta ABC$ with $AD$ as the internal angle bisector of $\angle A$. This means $\angle BAD=\angle CAD=\frac{A}{2}$.

Now, we draw a line through $D$ that is strictly perpendicular to $AD$, intersecting $AC$ at $E$ and $AB$ at $F$. This creates two right-angled triangles: $\Delta ADE$ and $\Delta ADF$.

Look at them closely. They share the side $AD$, they have equal angles at $A$, and they both have a $90^{\circ }$ angle at $D$. By the Angle-Side-Angle (ASA) congruence criterion, $\Delta ADE\cong \Delta ADF$.

This is our first breakthrough: because the triangles are congruent, their corresponding sides must be equal. Thus, $AE=AF$, which immediately proves that $\Delta AEF$ is an isosceles triangle. We have already conquered one of the options!

The Area Summation Principle

Now, let's tackle the length of the bisector $AD$. A common trap in JEE problems is to overcomplicate the algebra. Instead, let's use the Area Summation Principle.

The area of the large triangle $\Delta ABC$ is simply the sum of the areas of the two smaller triangles, $\Delta ABD$ and $\Delta ADC$. We know the area formula $\text{Area}=\frac{1}{2}ab\sin \theta$.

Applying this to our triangles, we get:

To simplify this, we use the double-angle identity: $\sin A=2\sin \frac{A}{2}\cos \frac{A}{2}$. Substituting this into our equation, we get:

Notice how the common term $\frac{1}{2}\sin \frac{A}{2}$ appears on both sides? We can cancel it out! This leaves us with:

This is the standard formula for the length of an internal angle bisector. We have just proven Option 2!

The Harmonic Mean and the Final Stretch

With $AD$ in our toolkit, finding $AE$ becomes a simple trigonometric exercise. In the right-angled $\Delta ADE$, we know that $\cos \frac{A}{2}=\frac{AD}{AE}$.

Rearranging this gives $AE=\frac{AD}{\cos \frac{A}{2}}$. Substituting our expression for $AD$, we get:

The $\cos \frac{A}{2}$ terms cancel out beautifully, leaving $AE=\frac{2bc}{b+c}$. This is the definition of the Harmonic Mean of $b$ and $c$. Option 1 is confirmed!

Finally, for $EF$, since $\Delta AEF$ is isosceles and $AD\perp EF$, $D$ is the midpoint of $EF$, so $EF=2\cdot DE$. In $\Delta ADE$, $\tan \frac{A}{2}=\frac{DE}{AD}$, so $DE=AD\tan \frac{A}{2}$.

Substituting $AD$ and using the identity $\cos \frac{A}{2}\tan \frac{A}{2}=\sin \frac{A}{2}$, we arrive at:

Option 3 is also correct! Isn't it satisfying when everything cancels out so perfectly? That is the beauty of geometry. Keep practicing, and you will start to see these symmetries everywhere.