Question 4 of 14advanced🔧 ApplyLong Answer3 marks

DD is a point on the side BCBC of a triangle ABCABC such that ∠ADC=∠BAC\angle ADC = \angle BAC. Show that CA2=CB⋅CDCA^2 = CB \cdot CD.

Correct Answer

In ΔADC\Delta ADC and ΔBAC\Delta BAC, we have:

∠ADC=∠BAC(Given)\angle ADC = \angle BAC \quad \text{(Given)}

∠ACD=∠BCA(Common angle)\angle ACD = \angle BCA \quad \text{(Common angle)}

Therefore, by AA similarity criterion, ΔADC∼ΔBAC\Delta ADC \sim \Delta BAC.

When two triangles are similar, their corresponding sides are in the same ratio.

So, we get:

CACB=CDCA\frac{CA}{CB} = \frac{CD}{CA}

Cross-multiplying:

CA2=CB⋅CDCA^2 = CB \cdot CD

∴ Hence proved.

Exercise: EXERCISE 6.3 | Q: 13 | (Chapter: Page 25)
For More Understanding

Explanation

This question tests the understanding of similarity of triangles. The key insight is recognizing that triangles ADCADC and BACBAC share a common angle at CC, and the given condition ∠ADC=∠BAC\angle ADC = \angle BAC provides the second angle equality.

By AA (Angle-Angle) similarity criterion, these triangles are similar. The context reinforces this approach through Example 3 which uses similarity to prove geometric relationships.

Once similarity is established, the property that corresponding sides of similar triangles are proportional is applied. The proportion CACB=CDCA\frac{CA}{CB} = \frac{CD}{CA} directly leads to CA2=CB⋅CDCA^2 = CB \cdot CD by cross-multiplication, which is the required result.

Solution Steps

  1. Step 1: Consider ΔADC\Delta ADC and ΔBAC\Delta BAC

  2. Step 2: Note that ∠ADC=∠BAC\angle ADC = \angle BAC (given) and ∠ACD=∠BCA\angle ACD = \angle BCA (common angle)

  3. Step 3: Apply AA similarity criterion to conclude ΔADC∼ΔBAC\Delta ADC \sim \Delta BAC

  4. Step 4: Write the proportion of corresponding sides:

    CACB=CDCA\frac{CA}{CB} = \frac{CD}{CA}

  5. Step 5: Cross-multiply to obtain:

    CA2=CB⋅CDCA^2 = CB \cdot CD

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