1. **Stating the problem:** We have a quadrilateral ABDM with diagonal AC. Angles at vertices are labeled as follows: angles $p$ and $w$ at vertex A, angle $t$ at vertex B, and angles 32° and 67° at vertex C, split by a line into parts $C$ and $K$. Sides AD and AC are equal, indicating triangle properties. 2. **Using triangle properties:** Since AD = AC, triangle ADC is isosceles with base DC. Therefore, angles opposite these equal sides are equal. Let angle $p$ and angle $w$ be the base angles at A. 3. **Finding $p$ and $w$:** In triangle ADC, the sum of angles is 180°: $$p + w + 32 + 67 = 180$$ Since 32° and 67° are angles at C split by the diagonal, their sum is 99°. So, $$p + w + 99 = 180$$ $$p + w = 81$$ Because AD = AC, $p = w$. Thus, $$2p = 81$$ $$p = w = \frac{81}{2} = 40.5$$ 4. **Finding $k$:** Angles at C are 32° and 67°, so $k$ corresponds to 67°. 5. **Finding $t$:** In triangle ABD, sum of angles is 180°. Angles at A are $p$ and $w$ (both 40.5°), and angle at B is $t$. Since $p$ and $w$ are at A, but the problem states $t$ is at B, and $p$ and $w$ are at A, we consider triangle ABD with angles $p$, $w$, and $t$. Sum of angles in triangle ABD: $$p + w + t = 180$$ $$40.5 + 40.5 + t = 180$$ $$81 + t = 180$$ $$t = 99$$ **Final answers:** $$w = 40.5^\circ$$ $$p = 40.5^\circ$$ $$k = 67^\circ$$ $$t = 99^\circ$$