Question

If one angle of a triangle is equal to one angle of the other triangle and the sides included between these angles are proportional then prove that the triangles are similar.

Hint:

The key to this proof is the construction step. By creating a triangle (\(\triangle DPQ\)) that is congruent to \(\triangle ABC\), you can transfer the properties of \(\triangle ABC\) into \(\triangle DEF\) and use theorems related to parallel lines to complete the proof.

Answer 1

Step 1: Understanding the Concept:
This is a proof of the Side-Angle-Side (SAS) similarity criterion for triangles. We need to prove that if two triangles have a pair of equal corresponding angles and the sides including these angles are in proportion, then the triangles are similar.

Step 2: Statement and Given Information:
Let the two triangles be \(\triangle ABC\) and \(\triangle DEF\).
Given: \(\angle A = \angle D\) and \(\frac{AB}{DE} = \frac{AC}{DF}\).
To Prove: \(\triangle ABC \sim \triangle DEF\).

Step 3: Construction and Proof:
Construction: On the side DE, cut a segment \(DP = AB\), and on the side DF, cut a segment \(DQ = AC\). Join PQ.
Proof:
In \(\triangle ABC\) and \(\triangle DPQ\): \[\begin{array}{rl} \bullet & \text{\(AB = DP\) (By construction)} \\ \bullet & \text{\(\angle A = \angle D\) (Given)} \\ \bullet & \text{\(AC = DQ\) (By construction)} \\ \end{array}\] Therefore, by SAS congruence rule, \(\triangle ABC \cong \triangle DPQ\).
This implies that \(\angle B = \angle DPQ\) and \(\angle C = \angle DQP\).
Now, it is given that \(\frac{AB}{DE} = \frac{AC}{DF}\).
Substituting \(AB = DP\) and \(AC = DQ\), we get: \[ \frac{DP}{DE} = \frac{DQ}{DF} \] By the converse of the Basic Proportionality Theorem (Thales's Theorem), this condition implies that \(PQ \parallel EF\).
Since \(PQ \parallel EF\), the corresponding angles are equal: \(\angle DPQ = \angle E\) and \(\angle DQP = \angle F\).
From our earlier findings, we know \(\angle B = \angle DPQ\) and \(\angle C = \angle DQP\).
Therefore, we can conclude that \(\angle B = \angle E\) and \(\angle C = \angle F\).
Now, in \(\triangle ABC\) and \(\triangle DEF\): \[\begin{array}{rl} \bullet & \text{\(\angle A = \angle D\) (Given)} \\ \bullet & \text{\(\angle B = \angle E\) (Proved above)} \\ \bullet & \text{\(\angle C = \angle F\) (Proved above)} \\ \end{array}\] Since all three corresponding angles are equal, by AAA similarity criterion, \(\triangle ABC \sim \triangle DEF\).

Step 4: Final Answer:
Hence, it is proved that the two triangles are similar.