Question

In the given figure, \(\Delta ABC\) is a right-angled triangle with \(\angle A = 90^\circ\). AD is perpendicular to BC.

34(a)(i) Prove that \(\Delta DBA \sim \Delta DAC\)

Hint:

An important theorem states: "If a perpendicular is drawn from the vertex of the right angle of a right triangle to the hypotenuse, then triangles on both sides of the perpendicular are similar to the whole triangle and to each other."
Remembering this theorem helps you predict the similarity relations instantly.

Answer 1

Correct Answer: 4

Step 1: Understanding the Question:
We are given \(\Delta ABC\) which is right-angled at \(A\) (\(\angle BAC = 90^\circ\)), with altitude \(AD \perp BC\).
We need to prove that the two smaller right-angled triangles formed, \(\Delta DBA\) and \(\Delta DAC\), are similar.

Step 2: Key Formula or Approach: We will use the Angle-Angle (AA) similarity criterion:
If two angles of one triangle are equal to two angles of another triangle, then the triangles are similar.

Step 3: Detailed Explanation:

• Let us establish the relationships between the angles of the triangles.
- Let \(\angle B = x\).
- In the right-angled triangle \(\Delta ABD\) (\(\angle ADB = 90^\circ\)):
\[ \angle BAD = 90^\circ - x \] - We are given that the total angle \(\angle BAC = 90^\circ\). Therefore:
\[ \angle DAC = \angle BAC - \angle BAD = 90^\circ - (90^\circ - x) = x \] - In the right-angled triangle \(\Delta ACD\) (\(\angle ADC = 90^\circ\)):
\[ \angle ACD = 90^\circ - \angle DAC = 90^\circ - x \]

• Now, let us compare the triangles \(\Delta DBA\) and \(\Delta DAC\):
- In \(\Delta DBA\) and \(\Delta DAC\):
\[ \angle BDA = \angle ADC = 90^\circ \text{ (Given that } AD \perp BC) \] \[ \angle ABD = \angle DAC = x \text{ (Proved above)} \] - Since two angles are equal, by the AA similarity criterion:
\[ \Delta DBA \sim \Delta DAC \]


Step 4: Final Answer:
The similarity \(\Delta DBA \sim \Delta DAC\) is proved.