Analysis: The only way to do this kind of problem is analysis, from unknown to known. To prove that EH is perpendicular to GH, then ∠EHG =∞.

Chg = 90. Because ch ⊥ AP = > ∠CHP=∠CHG+∠GHP=90。 Therefore, it is necessary to prove that ∠CHE=∠GHP, because AC is perpendicular to PD and CH is perpendicular to AP, so ∠ACH=∠CPA. In fact, we have to prove: △ car ∽△HPG. To prove that these two triangles are similar, only one set of diagonal is equal, and we just need to find the edge, so CH/PH=CE/PG must be established.

From known to unknown:

certificate

PB= 1/2AB, if p is the tangent of circle o, then: PDXPD=PBXPA=3POXPO/4. ∠CDP=90, we get: pd/po = ∠ 3/2. So: CPA = 30. ∠DCP =∞。 So ∠BEP=∠GPE, plus male ***∠PGA. Get △PFG∽△PAG, then, PGXPG=FGXAG. Because GD is tangent, so: GDXGD=FGXAG, so PG=DG. PB=BO=BD, so BG⊥PD. So, the quadrilateral BGCE is rectangular. So CE=BG. Because two right triangles, △PHC∽△PBG, CH/PH=BG/PG, because CE=BG. So CH/PH=CE/PG, so add: ∠ACH=∠CPA to get: △CHE∽△HPG, so ∠CHE=∠GHP because ch ⊥ AP = > ∠CHP=∠CHG+∠GHP=90。 So ∞∠EHG =∞.

Cut +∠CHG=∠CHG+∠GHP=90. Therefore, it is proved that EH is vertical to GH.

If you don't understand, you can ask questions, and some places are too simple to be omitted, such as projective theorem. . Equal circle angle and fox, tangent theorem.