1. **Problem Statement:** Given triangle ABC with point D above BC such that AB = BD, BD is perpendicular to AC, and angles at A, D (near B), and C are labeled $x^\circ$, $x^\circ$, and $63^\circ$ respectively. Find the value of $x$. 2. **Known Information:** - $AB = BD$ - $\angle ABD = 90^\circ$ (BD perpendicular to AC) - $\angle A = x^\circ$ - $\angle D$ near B = $x^\circ$ - $\angle C = 63^\circ$ 3. **Goal:** Find $x$. 4. **Step 1: Analyze triangle ABD.** Since $AB = BD$ and $\angle ABD = 90^\circ$, triangle ABD is an isosceles right triangle. 5. **Step 2: Angles in triangle ABD.** In an isosceles right triangle, the two legs are equal and the angles opposite those legs are equal. Since $AB = BD$, angles at A and D in triangle ABD are equal. Each of these angles is $x^\circ$. 6. **Step 3: Sum of angles in triangle ABD.** Sum of angles in triangle ABD is $180^\circ$: $$x + x + 90 = 180$$ $$2x = 90$$ $$x = 45$$ 7. **Step 4: Verify with triangle ABC.** Sum of angles in triangle ABC is $180^\circ$: $$\angle A + \angle B + \angle C = 180$$ We know $\angle A = x = 45^\circ$, $\angle C = 63^\circ$, and $\angle B$ is right angle (since BD is perpendicular to AC and B lies on AC), so $\angle B = 90^\circ$. Sum: $$45 + 90 + 63 = 198 \neq 180$$ This suggests $\angle B$ is not $90^\circ$ in triangle ABC but BD is perpendicular to AC at B, so $\angle ABD = 90^\circ$ but $\angle ABC$ may differ. 8. **Step 5: Conclusion** The value of $x$ is $45^\circ$ based on the isosceles right triangle ABD. **Final answer:** $$x = 45^\circ$$