Answer:
Hence proved △ABE∼△CBF.
Step-by-step explanation:
Given,
ABCD is a parallelogram.
BF ⊥ CD and
BE ⊥ AD
To Prove : △ABE∼△CBF
We have drawn the diagram for your reference.
Proof:
Since ABCD is a parallelogram,
So according to the property of parallelogram opposite angles are equal in measure.
⇒1
And given that BF ⊥ CD and BE ⊥ AD.
So we can say that;
⇒2
Now In △ABE and △CBF
∠A = ∠C (from 1)
∠E = ∠F (from 2)
So by A.A. similarity postulate;
△ABE∼△CBF
Answer:
Step-by-step explanation:
P = {x : x is a real number between 2 and 7}
{x: 3,4,5,6}
Q = {x : x is all rational number between 2 and 7}
{4 }
We know that P contains all the rational numbers between 2 and 7.
And P ∪ Q and Q ∪ P each of them contain all the real numbers which are between 2 and 7.
{3,4,5,6}
Here Q is the proper subset of P
P ∩ Q = Q ∩ P = Q= {4}
The answer is f12 submit 8765
You can write it as: 40,048,000,000
That answer is a 1 in 8 chance