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2 years ago

ABCD is a parallelogram. AB>AD. Prove: m∠ADB>m∠BDC. Show your work (statement reason maybe?)

Mathematics

2 answers:

Gwar [14]2 years ago

7 0

In the parallelogram ABCD, join BD.

Consider the triangle Δ ABD.

It is given that AB > AD.

Since, in a triangle, angle opposite to longer side is larger, we have,

∠ ADB > ∠ ABD. --- (1)

Also, AB || DC and BD is a transversal.

Therefore,

∠ ABD = ∠ BDC

Substitute in (1), we get,

∠ ADB > ∠ BDC.

prisoha [69]2 years ago

4 0

Answer:

Step-by-step explanation:

Given that ABCD is a parallelogram and AB >AD

To prove that $m∠ADB>m∠BDC.$

let us join BD.

Consider triangle ABD. AB and AD are two sides that AB>AD

By triangle theorem we get

$m∠ADB>m∠ABD.$

Since opposite sides are parallel, we find that AB and Dc are parallel lines with BD as transversal.

Hence $m∠ABD=m∠BDC.$ (alternate angles)

It follow that

$m∠ADB>m∠BDC.$

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2 years ago

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zhenek [66]

Answer:

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Step-by-step explanation:

Here the amount of incoming and outgoing of water are equal. Then the amount of water in the tank remain same = 10 liters.

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3 years ago

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Answer:

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