For reference: In the triangle $ABC$ inscribed in a circle of diameter $AD$. Calculate $HD$ since $H$ is the orthocenter and the distances from the circumcenter to sides $AB$ and $AC$ are $2.5$ and $1.5$ respectively with $\angle BAC=60^o$ (Answer:$\sqrt{19}$)
My progress and relationships I found:
$\triangle AOE \sim \triangle ADC \implies \frac{2R}{R} =\frac{AC}{AE} \therefore k = \frac{1}{2}\implies \\ DC = 2\cdot 1.5 = 3\\ Draw ~BD: \implies \triangle ABD \sim \triangle AGO\\ AO = AD \implies k = \frac{1}{2}\\ \therefore BD = 2\cdot OP = 2\cdot 2.5 = 5$
ABCD is cyclic
Draw $HD \rightarrow \triangle AHD\rightarrow HO$ is median $\implies$
T.Apollonius
$HD^2 +AH^2 = 2HO^2+\frac{R^2}{2}$
With MathLover hints: $\mathsf{\triangle ABD \sim \triangle APO \implies \frac{2R}{R} =\frac{BD}{PO} \therefore k = 2\\ \therefore BD = 2 \cdot OP = 2.2,5 = 5\\ DG \perp FB (G \in FB)\\ \angle GBD = 60^o \implies GD = \frac{5\sqrt3}{2}~e~BG = \frac{5}{2}\\ (By~property)BH=2OE \therefore BH = 3\\ GH = BH-BG=3-2,5 \implies GH=\frac{1}{2} \\ T.Pit: \triangle DHG:HD^2 = GH^2+GD^2 \implies\\ HD^2 = \frac{75}{4} +(\frac{1}{2})^2 \therefore \boxed{\color{red}HD = \sqrt19}}$