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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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Yes. No graph paper.

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Let's say the 3 points are A, B, C.

If A, B and C lie on one line then m_{AB} = m_{BC} = m_{AC}

m_{AB} = \frac{y_{B}-y_{A} }{x_{B}-x_{A}} = \frac{2-6}{3-5} = \frac{-4}{-2} = 2\\m_{BC} = \frac{y_{C}-y_{B} }{x_{C}-x_{B}} = \frac{8-2}{6-3} = \frac{6}{3} = 2\\\\m_{AC} = \frac{y_{C}-y_{A} }{x_{C}-x_{A}} = \frac{8-6}{6-5} = \frac{2}{1} = 2\\\\

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\boxed{\boxed{B.\ f(x)=\dfrac{3}{4}x^2+2x-5}}

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