Question

In the adjoining figure, $\triangle ABE \cong \triangle ACD$. Prove that (i) $\triangle ADE \sim \triangle ABC$ and (ii) $\triangle BOD \sim \triangle COE$.

Hint:

If two triangles are congruent, their corresponding ratios are 1:1, which automatically satisfies the side-proportionality requirement for similarity.

Answer 1

Step 1: Understanding the Concept:
Congruence ($\cong$) implies equal sides and angles. Similarity ($\sim$) requires proportional sides or equal angles.
Step 2: Key Formula or Approach:
1. CPCT (Corresponding Parts of Congruent Triangles).
2. SAS (Side-Angle-Side) Similarity Criterion.
Step 3: Detailed Explanation:
1. Part (i):
- Given $\triangle ABE \cong \triangle ACD$, so $AB = AC$ and $AE = AD$ (CPCT).
- In $\triangle ADE$ and $\triangle ABC$:
- $\frac{AD}{AB} = \frac{AE}{AC}$ (Since $AD=AE$ and $AB=AC$).
- $\angle A = \angle A$ (Common angle).
- Thus, $\triangle ADE \sim \triangle ABC$ by SAS similarity.
2. Part (ii):
- From $\triangle ABE \cong \triangle ACD$, we have $\angle ABE = \angle ACD$.
- Let intersection of $BE$ and $CD$ be $O$.
- In $\triangle BOD$ and $\triangle COE$:
- $\angle BOD = \angle COE$ (Vertically opposite angles).
- $\angle ODB = \angle OEC$ (Since $\triangle ADE \sim \triangle ABC$, $DE \parallel BC$, leading to equal alternate/corresponding angles).
- Thus, $\triangle BOD \sim \triangle COE$ by AA similarity.
Step 4: Final Answer:
Both similarities are proved using properties of congruence and parallel lines derived from the common vertex.