Triangle with \(BD=AD=AC\) and \( \angle DAC=96^\circ \) — Find \( \angle C \)

A crisp angle-chasing gem. The trick: use two isosceles triangles that share segment \(AD\) and the line \(BC\).

Concept in plain English

Point \(D\) lies on side \(BC\). We’re told three segments are equal: \(BD=AD=AC\). That means:

\(AD=AC\) ⇒ triangle \(ADC\) is isosceles with vertex at \(A\), so base angles at \(D\) and \(C\) are equal.
\(AD=BD\) ⇒ triangle \(ABD\) is isosceles with vertex at \(D\), so base angles at \(A\) and \(B\) are equal.
\(D\) lies on line \(BC\), so the ray \(CD\) and the ray \(CB\) are collinear (same line).

Show reference figure

Triangle with D on BC and equal segments — Given \(BD=AD=AC\) and \( \angle DAC=96^\circ \). Find \( \angle C=\angle ACB\).

CAT-style Question

Q. In triangle \( \triangle ABC \), a point \(D\) lies on \(BC\). If \(BD=AD=AC\) and \( \angle DAC=96^\circ\), find \( \angle C\).

Step-by-step solution (clean angle chase)

Work inside \( \triangle ADC\). Since \(AD=AC\), it’s isosceles with vertex angle \(\angle DAC=96^\circ\). Therefore the base angles at \(D\) and \(C\) are equal: \[ \angle ADC=\angle ACD=\frac{180^\circ-96^\circ}{2}=42^\circ. \]
Switch to \( \triangle ABD\). Since \(AD=BD\), base angles at \(A\) and \(B\) are equal: \[ \angle BAD=\angle ABD = x\ (\text{say}). \] The angle at \(D\) of \( \triangle ABD\) equals the angle between \(AD\) and \(DB\). Because \(DB\) is collinear with \(DC\), that angle is the supplement of \(\angle ADC\): \[ \angle ADB = 180^\circ - \angle ADC = 180^\circ - 42^\circ = 138^\circ. \] Sum of angles in \( \triangle ABD\): \(x+x+138^\circ=180^\circ \Rightarrow x=21^\circ\).
Find the big angle at \(A\). \(\angle BAC = \angle BAD + \angle DAC = 21^\circ + 96^\circ = 117^\circ.\)
Finish with triangle \(ABC\). Angles sum to \(180^\circ\): \[ \angle B + \angle C = 180^\circ - \angle A = 180^\circ - 117^\circ = 63^\circ. \] But \(\angle B\) contains \(\angle ABD = 21^\circ\) and a zero extra turn since \(DB\) and \(CB\) are the same line. Hence \(\angle B=21^\circ\). Therefore, \[ \angle C = 63^\circ - 21^\circ = \boxed{42^\circ}. \]

Show annotated solution figure

Angle chase with values filled in — Key values: \(\angle ADC=\angle ACD=42^\circ\), \(\angle BAD=\angle ABD=21^\circ\), \(\angle BAC=117^\circ\), so \(\angle C=42^\circ\).

Answer: \(\boxed{42^\circ}\)

Why this trick works (layman terms)

You create two isosceles triangles that touch at the same point \(D\). One (ADC) gives you two equal base angles of \(42^\circ\). The other (ABD) forces the angle at \(D\) to be the supplement of \(42^\circ\) (because the side \(DB\) sits on the same straight line as \(DC\)). That makes the two base angles \(21^\circ\) each, and the rest is just adding to \(180^\circ\).

CAT Geometry — When BD = AD = AC & angle DAC = 96°