As shown in the figure, ad is the middle line of △ ABC, be intersects AC to e, and ad to F, and AE = EF. Verification: AC = BF

In △ BDF and △ CDM, BD = CD, BD = CD. In △ BDF and △ CDM, BD = CD, BDF = cdmdf = DM ≌ BDF ≌△ CDM (SAS)

In the triangle row ABC, CE is perpendicular to AB and BF is perpendicular to ac. it is proved that the AEF of triangle is similar to that of ACB a = 60 degrees, and the area ratio of AFE ABC is obtained

There are BF vertical AC, CE vertical AB, ∠ a = ∠ a, so there is triangle AFB which is similar to triangle AEC
Therefore, triangle AEF is similar to triangle ACB
When ∠ a = 60, ∠ ABF = 30 degrees, AE = 1 / 2Ab, that is, the similarity ratio is 1 / 2, and the area ratio is 1 / 4 of the square of the similarity ratio

It is known that CE in triangle ABC is perpendicular to AB and BF is perpendicular to AC and F. it is proved that the triangle AFE is similar to triangle ABC

In △ AEC and △ AFB
∵∠A=∠A,∠AEC=∠AFB
∴△AEC∽△AFB
∴AE:AF=AC:AB
For △ AEF and △ ACB
∵∠A=∠A,AE:AF=AC:AB
∴△AFE∽△ABC

It is known that in the triangle ABC, CE ⊥ AB is in E, BF ⊥ AC is in F. if s ⊥ ABC = 36 and s △ AEF = 4, find the value of sina Triangle ABC is an acute triangle

Because the angle AEC = angle AFB = 90 degrees, angle a = angle a, the triangle AEC is similar to the triangle AFB, so AE / AF = AC / ab. according to this condition, plus the common angle a, the triangle AEF is similar to the triangle ACB, because the area ratio is equal to the square of the side length ratio, 36 / 4 = 9 / 1 = (3 / 1), so AE / AC = AF / AB =

In the triangle ABC, CE is perpendicular to AB and BF is perpendicular to ac. it is proved that △ AEF is similar to △ ACB

Because, ∠ BEC = 90 ° = ∠ BFC, so B, e, F, C are in the same circle, we can get: ∠ AFE = ∠ ABC, and ∠ A is the common angle of △ AEF and △ ABC. Therefore, △ AEF ∽ △ ABC. If you have not learned the circle, you can use the following method: in △ ace and △ ABF, ∠ AEC = 90 ° = ∠ AFB, ∠ A is

In the triangle ABC, ab = AC, a point on the edge of D bit BC, BF = CD, CE = BD, find the degree of angle EDF It's the relationship between angle EDF and angle A

If AB = AC, then angle B = angle C
Because BF = CD, CE = BD, then FBD of triangle is equal to DCE of triangle, then angle DFB = angle EDC
Because angle DFB + angle B = angle FDC = angle FDE + angle EDC
Then angle B = angle FDE
Because the angle B + angle a is 180
Then the EDF + a = 180

As shown in the figure, in the triangle ABC, ab = AC, BF = CD, BD = CE, ∠ FDE = α, then the relationship between α and ∠ A is

Junlang lieying team answers for you
∵AB=AC,∴∠B=∠C,
∵BF=CD,BD=CE,
∴ΔBDF≌ΔCED,
∴∠BFD=∠CDE,
∵∠B+∠BFD+∠BDF=180°,
∠BDF+α+∠CDE=180°,
∴∠B=α.
And ∠ a = 180 ° - 2 ∠ B,
∴∠A=180°-2α.

In the p-abcd of a pyramid, ∠ ABC = ∠ ACD = 90 °, BAC = ∠ CAD = 60 ° PA ⊥ plane ABCD, e is the midpoint of PD, PA = 2Ab = 2 (I) if f is the midpoint of PC, verify the AEF of PC ⊥ plane; (II) confirm CE ‖ plane PAB

In the pyramid p-abcd, the angle ABC = angle ACD = 90 ° angle BAC = angle CAD = 60 ° PA vertical plane ABCD, e is the midpoint of PD, PA = 2Ab = 2 (1) To find the volume V of p-abcd (2) If f is the midpoint of PC: plane PAC vertical plane AEF (3) Finding dihedral angle e-ac-d

(1) V = 1 / 3 × PA × s bottom
S base = s △ ABC + s △ ACD = 1 / 2 × 1 × root 3 + 1 / 2 × 2 × 2 times root 3 = 5 / 2 times root sign 3
Therefore, v = 1 / 3 × 2 × 2 / 5 times root sign 3 = 5 / 3 times root sign 3
(2) Link ef
Because e and F are the midpoint of PD and PC respectively
So EF is parallel and equal to 1 / 2CD
Because the ABCD of PA ⊥ surface is included in the surface ABCD
So EF ⊥ pa
And ∠ ACD = 90 ° DC ⊥ AC
So EF ⊥ AC
PA∩AC=A
So EF ⊥ PAC
EF included in face AEF
So PAC ⊥ AEF
Wait a minute. I'll have a meal to replenish my energy

In p-abcd of tetrahedral pyramid, ∠ ABC = ∠ ACD = 90 °, BAC = ∠ CAD = 60 ° and E is the midpoint of PD to verify CE ‖ plane PAB

Take the midpoint F of AD, connect CF and EF, ∵ EF is the median line of △ pad,

As shown in the figure, ad is the middle line of △ ABC, be intersects AC to e, and ad to F, and AE = EF. Verification: AC = BF