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Q7(iv):

In Fig. 6.38, altitudes $AD$ and $CE$ of $\triangle ABC$ intersect each other at the point $P$. Show that: (iv) $\triangle PDC \sim \triangle BEC$

Solution :

Given: In $\triangle ABC$, $AD \perp BC$ and $CE \perp AB$. The altitudes $AD$ and $CE$ intersect each other at point $P$.

To Prove: $\triangle PDC \sim \triangle BEC$.

Visual Representation:

B C A D E P

Step 1: Identify the triangles to be compared.
We are considering $\triangle PDC$ and $\triangle BEC$.

Step 2: Analyze the angles of $\triangle PDC$ and $\triangle BEC$.
In $\triangle PDC$ and $\triangle BEC$:

1. Consider $\angle PCD$ and $\angle BCE$.
Since $P$ lies on $CE$ and $D$ lies on $BC$, the angle $\angle PCD$ is the same as $\angle BCE$.
Therefore, $\angle PCD = \angle BCE$ [Common angle].

2. Consider the right angles in both triangles:
In $\triangle PDC$, $AD \perp BC$, therefore $\angle PDC = 90^\circ$.
In $\triangle BEC$, $CE \perp AB$, therefore $\angle BEC = 90^\circ$.
Thus, $\angle PDC = \angle BEC = 90^\circ$ [Given that $AD$ and $CE$ are altitudes].

Step 3: Apply the AA (Angle-Angle) Similarity Criterion.
The AA Similarity Criterion states that if two angles of one triangle are respectively equal to two angles of another triangle, then the two triangles are similar.
From Step 2, we have:
$\angle PCD = \angle BCE$ (Common)
$\angle PDC = \angle BEC$ (Each is $90^\circ$)

Step 4: Conclusion of Similarity.
By the AA similarity criterion, since two corresponding angles are equal, the triangles must be similar.
$\triangle PDC \sim \triangle BEC$ [By AA Similarity Criterion].

Final Answer: Since $\angle PCD = \angle BCE$ and $\angle PDC = \angle BEC = 90^\circ$, by the AA similarity criterion, $\triangle PDC \sim \triangle BEC$.


More Questions from Class 10 Mathematics Triangles EXERCISE 6.3


CBSE Solutions for Class 10 Mathematics Triangles


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