<h2><em><u>REFERS </u></em><em><u>TO </u></em><em><u>THE </u></em><em><u>ATTACHMENT</u></em><em><u>.</u></em></h2>
<h2><em><u>PLEASE</u></em><em><u> </u></em><em><u>MARK</u></em><em><u> ME</u></em><em><u> AS</u></em><em><u> BRAINLIEST</u></em><em><u>:</u></em></h2>
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
well, it's isosceles so use base angles theorem
the top angle is also x
90 + 2x = 180
subtract 90 from both sides
2x = 90
divide both sides by 2
x = 45 degrees
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Measurement of "AC" :
(x + 5) + (2x <span>− 11) ;
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Find the measurement of "AB" [which is: "(x+5)" ]:
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First, simplify to find the measurement of "AC" :
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</span>(x + 5) + (2x − 11) ;
= (x + 5) + 1(2x − 11) ;
= x + 5 + 2x − 11 ;
→ Combine the "like terms" ;
x + 2x = 3x ;
5 − 11 = - 6 ;
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to get: 3x − 6 ;
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So, (x + 5) + (2x − 11) = 3x − 6 ;
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Solve for: "(x + 5)"
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We have:
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(x + 5) + (2x − 11) = 3x − 6 ;
Subtract: "(2x − 11)" ; from EACH SIDE of the equation ;
to isolate "(x + 5)" on one side of the equation;
and to solve for "(x + 5)" ;
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→ (x + 5) + (2x − 11) − (2x − 11) = (3x − 6) − (2x − 11) ;
→ (x + 5) = (3x − 6) − (2x − 11) ;
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Note: Simplify: "(3x − 6) − (2x − 11)" ;
→ (3x − 6) − (2x − 11) ;
= (3x − 6) − 1(2x − 11) ;
= 3x − 6 − 2x + 11 ;
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→ Combine the "like terms" :
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+3x − 2x = 1x = x ;
-6 + 11 = 5 ;
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To get: x + 5 ;
So we have:
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x + 5 = x + 5 ;
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So, x = all real numbers.
x = <span>ℝ </span>