Show that if two such isosceles triangles (occurring in the previous question) have equal base length, they are congruent to each other.

Question

Homi · Accepted Answer

Understand the Question

Two isosceles triangles are formed in a circle with the center as the common vertex, where the two equal legs are radii of the circle and the base is a chord.
Since all radii of a circle are equal, both pairs of legs in the two triangles are equal.
Given that their bases (chords) are also equal, all three corresponding sides of the two triangles are equal.
Therefore, the triangles are congruent by the SSS (Side-Side-Side) congruence criterion.

Step 1 · Identify the Triangles and Given Information

Let the two isosceles triangles in the circle with center $O$ be $\triangle OAB$ and $\triangle OPQ$ , where chords $AB$ and $PQ$ are the respective bases. Diagram 1

Given that the base lengths are equal: $AB = PQ$

Step 2 · Apply SSS Congruence Criterion

In $\triangle OAB$ and $\triangle OPQ$ :

\begin{aligned} OA &= OP \quad (\text{Radii of the same circle}) \\ OB &= OQ \quad (\text{Radii of the same circle}) \\ AB &= PQ \quad (\text{Given equal base length}) \end{aligned}

By the Side-Side-Side (SSS) congruence criterion: $\triangle OAB \cong \triangle OPQ$

Answer

$\triangle OAB \cong \triangle OPQ$ (by SSS congruence criterion)

Common Mistakes

Assuming Angle Equality Directly: Trying to use SAS criterion by assuming the central angles $\angle AOB = \angle POQ$ without proving it first. Since only side lengths are provided, SSS is the direct and correct criterion to use.
Overlooking Radii Equality: Forgetting that all radii in a circle are equal, which directly gives $OA = OB = OP = OQ$ .

Question 2

More questions in Exercise 5.2