Question

can $\triangle abd$ be proved to be congruent to $\triangle cdb$ by the angle-side-angle triangle congruency theorem? in the figure, $\angle adc \cong \angle cba$ and $\angle adb \cong \angle cbd$.

figure of a quadrilateral with triangles abd and cdb, and angles marked

\bigcirc yes
\bigcirc no

explain or show your reasoning.

Explanation:

Step1: Analyze Given Angles

We know \( \angle ADC \cong \angle CBA \) and \( \angle ADB \cong \angle CBD \). Let's find the relationship between \( \angle ABD \) and \( \angle CDB \). Since \( \angle ADC=\angle ADB + \angle CDB \) and \( \angle CBA=\angle ABD+\angle CBD \), and \( \angle ADC\cong\angle CBA \), \( \angle ADB\cong\angle CBD \), by angle subtraction (if \( \angle ADC=\angle ADB+\angle CDB \) and \( \angle CBA=\angle ABD + \angle CBD \), then \( \angle ADB+\angle CDB=\angle ABD+\angle CBD \), and since \( \angle ADB = \angle CBD \), we get \( \angle CDB=\angle ABD \)).

Step2: Identify Common Side

The side \( BD \) is common to both \( \triangle ABD \) and \( \triangle CDB \), so \( BD\cong BD \) (reflexive property).

Step3: Apply ASA Theorem

ASA (Angle - Side - Angle) requires two angles and the included side. We have \( \angle ADB\cong\angle CBD \), \( BD\cong BD \), and \( \angle ABD\cong\angle CDB \). So in \( \triangle ABD \) and \( \triangle CDB \), \( \angle ADB=\angle CBD \), \( BD = BD \), \( \angle ABD=\angle CDB \), which satisfies the ASA congruence condition (the side \( BD \) is included between the two angles in each triangle).

Answer:

Yes