30 points for junior high school math proof questions! A junior high school math proof problem! Urge

First prove that it is a parallelogram, like that person, and because ab=3, ac=4, bc=5, we can see that abc is a right triangle, and bac is a right angle.

Then calculate the degree of the angle EAD, which should be equal to 360-90-60-60=150°, then according to the quadrilateral theorem, we can know that ADF=30°, and then make a perpendicular line from F to AD, and the vertical foot is G, because FD=AE=AC=4, so we can know FG=2, and because AD=AB=3

So the area of the quadrilateral AEFD is 3*2=6

The point A is perpendicular to EF, the quadrilateral AEFD is a parallelogram (which can be proved by that question), AD=AB=4, BAC=90°(3,4,5 triangles), and the inner angle of the regular triangle is 60°, so HAE=60°

ae=ac=3, so ah=3 2, the area of the quadrilateral AEFD is 4*(3 2)=6

You can do the math yourself, right?

First of all, I would like to say a few things.

It should be ge=ec, right?

Then the answer is as follows:

Angular AEG = Angular EAC + Angular ACE

Angular agb = angular gae + angular aeg = angular gae + angular eac + angular ace = angular gac + angular ace

Evidence: AGB = AEB + ACB is evidence that the angle GAC = angle AEC is similar with a triangle.

That is, AEG CAG

ae: ca = root 5: root 10

ag:cg=root2:2

eg:ag=1:root2

So similar. So angular GAC = angular AEC gets what you want to prove

Hello, I sincerely serve you.

In Sakura Jane High ABC.

Because aed= b+ 1 (

The outer corners of the triangle are noisy.

Equal to two that are not adjacent.

Inner corners. sum).

1= b, so the spine ruler aed=2 b

Therefore aed= c

And because AD is.

Angular bisector. ad is a common edge.

So triangle ACD

All equals. Triangular ade

aas) so ac=ae

cd=de, 1= b

So de=be

So. ab=ac+cd hope.

In junior high school, you may not know the cosine theorem, so make it into two right triangles, if you know the cosine theorem, it will be very easy......

Can you use the method of similar triangles? If I could, I'd do it above.