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: See the attached image
You have the correct idea for the boxes you've filled out. For the first three boxes in column 1, I would be specific which segments you are dividing. So for instance, in the first box, it would be EG/EB = 55/11 = 5. Then the second box would be EF/EC = 35/7 = 5, and so on. The order of the boxes doesn't matter. The three boxes then combine together to help show that the triangles are similar. Specifically . The order of the letters is important to help show how the angles pair up and how the sides pair up. We use the SSS similarity theorem here.
The second problem is the same idea, but we use one pair of congruent angles. So we'll use the SAS similarity theorem this time.
