ABC is a triangle, P is a point on AB, and Q is point on AC such that ∠AQP = ∠ABC. Complete the rela

Q: Wait, why can't I just assume $PQ$ is parallel to $BC$?

That is the classic trap! If $PQ$ were parallel, then $\angle APQ$ would equal $\angle ACB$. But the problem gives us $\angle AQP = \angle ABC$. This is a subtle 'anti-parallel' configuration. Never trust your eyes in geometry; trust the axioms! The math dictates the shape, not the drawing.

Q: Why does the order of vertices matter so much in $\Delta APQ \sim \Delta ACB$?

Because similarity is a mapping! If you write $\Delta APQ \sim \Delta ABC$, you are claiming $AP/AB = AQ/AC$. But our similarity is $\Delta APQ \sim \Delta ACB$, which implies $AP/AC = AQ/AB$. Mixing these up will invert your final answer. Always map the equal angles to each other.

Q: Is there a way to solve this using the sine rule for area?

Absolutely! $\text{Area} = \frac{1}{2} ab \sin C$. You could write the ratio of areas using $\frac{1}{2} AP \cdot AQ \sin A$ and $\frac{1}{2} AB \cdot AC \sin A$. It works beautifully, but similarity is the 'elegant' path that builds your geometric intuition for more complex JEE problems.

Question

$ABC$ is a triangle, $P$ is a point on $AB$, and $Q$ is point on $AC$ such that $\angle AQP = \angle ABC$. Complete the relation $\frac{	ext{area of } \Delta APQ}{	ext{area of } \Delta ABC} = \frac{(\dots)}{AC^2}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing before a triangle, $\Delta ABC$. You see a line segment $PQ$ cutting across it. Your intuition, honed by years of seeing standard diagrams, might scream, "Oh, $PQ$ is parallel to $BC$!"

Stop. Take a breath. In the world of competitive exams, your intuition is often the first trap.

The problem gives us a specific condition: $\angle AQP = \angle ABC$. This is not a parallel line configuration; it is an **anti-parallel** one. We must rely solely on the logic of angles, not the appearance of the sketch.

### The Hunt for Similarity

To find the ratio of the areas, we need to understand how $\Delta APQ$ relates to $\Delta ABC$. The most powerful tool in our arsenal is the **AA (Angle-Angle) Similarity Criterion**.

We already have one angle given: $\angle AQP = \angle ABC$. Now, look at the vertex $A$. It is the common vertex for both the small triangle $\Delta APQ$ and the large triangle $\Delta ACB$.

Therefore, $\angle PAQ = \angle CAB$. We have found our two pairs of equal angles! By the AA Similarity Criterion, we can confidently state that $\Delta APQ \sim \Delta ACB$.

### The Vertex Correspondence

This is where most students stumble. When we write $\Delta APQ \sim \Delta ACB$, the order of the letters is not arbitrary; it is a sacred map.

Vertex $A$ corresponds to $A$. Vertex $Q$ corresponds to $B$. Vertex $P$ corresponds to $C$. If you get this order wrong, your ratios will be inverted, and your answer will be incorrect.

Because the triangles are similar, the ratio of their corresponding sides must be equal. This gives us the relationship:

$$
\frac{AP}{AC} = \frac{AQ}{AB} = \frac{PQ}{CB}
$$

Notice that $AP$ corresponds to $AC$, not $AB$. This is the crux of the problem.

### The Area Ratio Theorem

Now, we arrive at the grand finale. There is a fundamental theorem in geometry: the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

Mathematically, this is expressed as:

$$
\frac{	ext{Area}(\Delta APQ)}{	ext{Area}(\Delta ACB)} = \left(\frac{AP}{AC}ight)^2
$$

When we expand this, we get:

$$
\frac{	ext{Area}(\Delta APQ)}{	ext{Area}(\Delta ACB)} = \frac{AP^2}{AC^2}
$$

The question asks us to complete the relation $\frac{	ext{area of } \Delta APQ}{	ext{area of } \Delta ABC} = \frac{(\dots)}{AC^2}$. By comparing our derived formula to the given expression, the answer reveals itself clearly: the missing term is **$AP^2$**.

You have successfully navigated the trap, respected the vertex correspondence, and applied the theorem with precision. This is the essence of JEE mastery—not just finding the answer, but understanding the path that leads there.

Animated Solution for Mathematics - Trigonometry: $A B C$ is a triangle, $P$ is a point on $A B$ , and $Q$ is point on $A C$ such that $∠ A QP = ∠ A B C$ . Complete the relation $\frac{area of Δ A P Q}{area of Δ A B C} = \frac{( \dots )}{A C ^{2}}$

Visualized Solution

Visualizing Triangle $A B C$ and Segment $P Q$

Identifying the Given Equal Angles

Finding the Common Angle

Establishing Similarity by AA Criterion

Mapping Corresponding Vertices

Writing the Ratio of Corresponding Sides

Applying the Area Ratio Theorem

Completing the Relation

The Sigma Insight: Properties of Triangles

Analyzing the Setup

The Hunt for Similarity

The Vertex Correspondence

The Area Ratio Theorem

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