As shown in the figure, it is known that the straight line ef parallel to AB CD and AB CD intersect at G point respectively, and HP is any point in HD. P doing a straight line pm meets EF at M. Explanation ∠hmp=∠agf-∠hpm Question Supplement: Can I help him answer it?

Satisfied answer 20 12-04-2 1? 19:58 Hello ∵AB is perpendicular to AC, ∴∠ BAC = 90 (vertical definition).

∠∠DAB =∠C (known), and∠∠∠ B =∠ B (male * * * angle)

∴∠ ∴∠BDA =∠BAC = 90°∴∠CDA = 90° (right angle definition)

Think of this problem this way: look at two triangles △ABC and△ △ADB, and we know that the corresponding two angles are equal, and the third angle is naturally equal. Because △ cab = 90 and △ ADB = 90, the problem is solved.

Remember, this figure is called double vertical figure, which is often used when learning similar triangles in grade three.

I can't do the second question without a picture ... Give a picture. ......