In the parallelogram ABCD, there is a point E on the side of DC, which connects AE. The point F is on the side of AE, and connects B. It is proved that the triangle ABF is similar to the triangle EAD Angle BFE is equal to angle C

In the parallelogram ABCD, there is a point E on the side of DC, which connects AE. The point F is on the side of AE, and connects B. It is proved that the triangle ABF is similar to the triangle EAD Angle BFE is equal to angle C

AB‖CD
There are: ∠ DEA = ∠ fab
∠ADE=180-∠C,∠AFB=180-∠EFB,
Because ∠ EFB = ∠ C
Therefore, ADE = AFB
So: Triangle ABF is similar to triangle EAD

In square ABCD, e, f feed the midpoint of AB, BC. DF, CF intersect m, proving am = ad
P16

According to the following conditions: △ BCE ≌ △ CDF (s, a, s),
∴∠BCE=∠CDF,
And ∠ BCE + ∠ DCE = 90 °,
∴∠CDF+∠DCE=90°,
∴CE⊥DF.
After a, AP ⊥ DM was transferred to P,
△ADP≌△DCM（A,S,A)
∴DM=AP,
Also △ DCM ∽ CEB (key)
∴CM：DM=1：2,
From CM = DP, DP = MP,
That is, P is the midpoint of DM,
Am = ad

In the parallelogram ABCD, M is the midpoint of BC, and am = 9, ad = 10, BD = 12, am and BD intersect at P,
To find the area of a parallelogram, please write down the process,

The parallel line of BD through M intersects the extension line of ad at n
In △ amn, an = 10 + 10 / 2 = 15, am = 9, Mn = BD = 12
It can be seen that the angle amn is a right angle. The height of its hypotenuse (that is, the height of the parallelogram row) is 9 * 12 / 15
So the area of parallelogram is: 10 * 9 * 12 / 15 = 72