Ask all the masters to solve the trapezoidal similar triangle puzzle

Because. The area ratio of the triangle AOD and the triangle ACD is 1:3, which means that the ratio of the height of the two triangles is 1:3

You can make perpendicular lines to AD and BC respectively through the O point, and set Oe to the height of the triangle AOD.

DF is the height of a triangular ADC.

It's oe:df 1:3

Then pass the O point to the BC lead line OG

So oe:og=1:2

The triangle OED is similar to the triangle OGB.

So ed:bg=1:2

Similarly. The triangular OAE and the triangular OCG are similar.

So ae:cg=1:2

So ad:bc=1:2

The height of the triangle AOD and the triangle COB.

oe：og=1：2

So. The area ratio of these two triangles is.

S-triangle aod: s-triangle cob=1:4

Not sure, yes! ^_

1:4 According to the title, the triangle AOD and the triangle ADC have the same bottom, so the ratio of their bases is 1:3

So ao:ac=1:3, then ao:

oc=ao:ac-ao=1:2, and then the triangle aod similar triangle boc is obtained from ad parallel bc, so their similarity ratio is 1:

2. The area ratio is 1:4So their s-triangle aod:

The ratio of the S-triangle BOC is 1:4

First of all, AOB is similar to COD, and AOD is equal to BOC area = 12. Let the upper bottom be a, the similarity ratio is k, and the upper height is h, then the lower bottom is ak, and the lower height is hk, ah 2+ak*hk 2=25(a+ak)*(h+hk) 2=49 to get k=3 4 or k=4 3

This question should be made with two auxiliary lines: 1. Take ab as the base, pass the point e as the high, and hand AB at g; 2. Take AB as the base, pass the C point as a high, and hand AB at H.

The triangle age is similar to the triangle AHC, then: eg ch=ag ah, where, eg=fb, ch=db, ag=, ah=ab-3, you can calculate the value of ab.

Solution: If the point c is parallel to db and ab is in g, then ag:cg=(:2=, so.

ag=So the flagpole height is 3+

The 3% of the first figure refers to the value of A to B as 3%. Then extend the b side of the second image and extend the hypotenuse to get the same triangle as the first image.