In △ ABC, ∠ C = 60 ° and high be passes through the midpoint F of high ad, be = 10cm, the length of BF and EF can be calculated

In △ ABC, ∠ C = 60 ° and high be passes through the midpoint F of high ad, be = 10cm, the length of BF and EF can be calculated

In the right angle △ BCE, ∠ C = 60 °
∴∠CBE=30°
∴BF=2DF
In the right angle △ ACD, ∠ C = 60 °
∴∠CAD=30°
∴EF=AF/2
AF = DF, be = 10
∴EF=2,BF=8

If be = 10 cm, the length of BF and EF can be calculated

In the right angle △ BCE, ∠ C = 60 °
∴∠CBE=30°
∴BF=2DF
In the right angle △ ACD, ∠ C = 60 °
∴∠CAD=30°
∴EF=AF/2
AF = DF, be = 10
∴EF=2,BF=8

As shown in the figure, in △ ABC, ∠ C = 90 ° and high be passes through the midpoint F of high ad. if be = 10cm, calculate the length of BF and ef
Wrong, ∠ C = 60 °
Ad is the height of BC, be is the height of AC (I can't draw)
I have the answer to this question, but I don't know how
BF=8，EF=2

Show me a picture

The two sides of an isosceles triangle are 6cm.12cm, and its circumference is (). Options: 1.24cm, 2.30cm, 3.24cm or 30cm

1198913536 ,
Perimeter: 12 × 2 + 6 = 30 (CM)
Choose 2.30 cm
It can't be 24 cm. When the other side is 6 cm, because 6 + 6 = 12 violates that the sum of any two sides is greater than the third side, it's impossible