In the figure shown, ABCD is a parallelogram. It is divided into triangles in such a way that $∠$ BEC = $∠$ BCE. Which one of the following is incorrect?

1, 2, 3

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1, 3

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Only 3

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Only 2

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Solution

The correct option is B
1, 3

Since $∠$ BEC = $∠$ BCE, $Δ$ BEC is an isosceles triangle.

So, BE = BC

AF is the shortest distance between point A and CD. As we move along CD towards D, E, and C, the distance from point A increases.

This means that AE > AD. So, AD > AE is incorrect

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In the figure shown, ABCD is a parallelogram. It is divided into triangles in such a way that ∠BEC = ∠BCE. Which one of the following is incorrect?

The correct option is B 1, 3 Since ∠BEC = ∠BCE, ΔBEC is an isosceles triangle. So, BE = BC AF is the shortest distance between point A and CD. As we move along CD towards D, E, and C, the distance from point A increases. This means that AE > AD. So, AD > AE is incorrect

In the figure shown, ABCD is a parallelogram. It is divided into triangles in such a way that $∠$ BEC = $∠$ BCE. Which one of the following is incorrect?

The correct option is B
1, 3

Since $∠$ BEC = $∠$ BCE, $Δ$ BEC is an isosceles triangle.

So, BE = BC

AF is the shortest distance between point A and CD. As we move along CD towards D, E, and C, the distance from point A increases.

This means that AE > AD. So, AD > AE is incorrect