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bonufazy [111]
2 years ago
12

Given: BCDF is a parallelogram; AB = BF Prove: angleA=angleE

Mathematics
1 answer:
Katen [24]2 years ago
7 0

Answer:

Step-by-step explanation:

ABCD is a parallelogram.  AB is produced to P, such that AB = BP and PQ is drawn parallel to BC to meet AC produced at Q. Given AB = 8 cm, AD = 5 cm, AC = 10 cm.

→ BC ║ QP →→Given

In Δ ABC and ΔAPQ

∠ABC=∠APQ→→[BC ║ QP , BP is a transversal, so corresponding angles are equal]

∠BAC=∠PAQ→→Reflex angle

Δ ABC ~ ΔAPQ→→(AA similarity criterion]

When triangles are similar, their corresponding angles are equal.

AC=CQ=10 cm, shows that point C is mid point of AQ.

<u>help me by marking as brainliest....</u>

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Answer:

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General Formulas and Concepts:

<u>Algebra I</u>

  • Solving Exponential Equations
  • Exponential Property [Rewrite]: \displaystyle b^{-m} = \frac{1}{b^m}

Step-by-step explanation:

<u>Step 1: Define</u>

<em>Identify</em>

<em />\displaystyle 5^a = \frac{1}{125}<em />

<em />

<u>Step 2: Solve for </u><em><u>a</u></em>

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