Question

In the given figure, DEFG is a square. \(\Delta ABC\) is right angle triangle with \(\angle A = 90^\circ\). Prove that AG \(\times\) DG = AF \(\times\) DB.

Hint:

To write corresponding side ratios correctly, always trace the order of vertices from the similarity statement:
If \(\Delta \textbf{A}\textbf{G}F \sim \Delta \textbf{D}\textbf{B}G\), then the first two letters give \(\frac{AG}{DB}\), and the first and third letters give \(\frac{AF}{DG}\).

Answer 1

Step 1: Understanding the Question:
The topic of this question is Triangles, specifically Similarity of Triangles.
We are given a right-angled triangle \(\Delta ABC\) where \(\angle A = 90^\circ\).
Inside it, there is a square \(DEFG\) with vertices \(D\) and \(E\) lying on the hypotenuse \(BC\), vertex \(G\) on \(AB\), and vertex \(F\) on \(AC\).
We need to prove that \(AG \times DG = AF \times DB\).

Step 2: Key Formula or Approach:
We will find similar triangles in the diagram:
- Since \(DEFG\) is a square, its opposite sides are parallel. Thus, \(GF \parallel BC\).
- Since \(GF \parallel BC\), the corresponding angles are equal:
\[ \angle AGF = \angle B \quad \text{and} \quad \angle AFG = \angle C \] - We will prove that \(\Delta AGF \sim \Delta DBG\) using the AA (Angle-Angle) similarity criterion.
- Once similarity is established, we will set up the ratio of corresponding sides and cross-multiply to prove the relation.

Step 3: Detailed Explanation:

• Establish the right angles:
Since \(DEFG\) is a square, the angle \(\angle GDB = 90^\circ\) because \(GD \perp BC\).
We are given that \(\angle A = 90^\circ\).
Thus:
\[ \angle A = \angle GDB = 90^\circ \]

• Relate the other angles:
In right-angled triangle \(\Delta ABC\):
\[ \angle B + \angle C = 90^\circ \] In right-angled triangle \(\Delta GDB\):
\[ \angle B + \angle BGD = 90^\circ \implies \angle BGD = 90^\circ - \angle B \] Since \(\angle B = \angle AGF\) (corresponding angles), and \(\angle AGF + \angle AFG = 90^\circ \implies \angle AFG = 90^\circ - \angle AGF\), we have:
\[ \angle AFG = \angle BGD \]

• Apply the AA similarity criterion to \(\Delta AGF\) and \(\Delta DBG\):
- \(\angle A = \angle GDB = 90^\circ\)
- \(\angle AFG = \angle BGD\)
Therefore:
\[ \Delta AGF \sim \Delta DBG \]

• Set up the ratio of corresponding sides from the similarity statement:
\[ \frac{AG}{DB} = \frac{AF}{DG} \]

• Cross-multiply the terms of the proportion:
\[ AG \times DG = AF \times DB \] This completes our geometric proof.


Step 4: Final Answer:
Hence, it is proved that AG \(\times\) DG = AF \(\times\) DB.