Question

a straight line parallel to ab of parallelogram ABCD intersects ad ,ac, BC at e, f , g respectively. proof that triangle AEG= triangle afd

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Answer 1

Answer:

AD + BC ) : AD

Step-by-step explanation:

Let h = height of trapezium and AD = a, BC = b

Area of ABCD = \frac{1}{2}

2

1

× h × (a + b)

Area of ΔAFD = \frac{1}{2}

2

1

× a × h

Ratio of area of ABCD : AFD = (a+b) : a

= (AD + BC ) : AD

Step-by-step explanation:

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