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natima [27]
3 years ago
15

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

Mathematics
2 answers:
Gwar [14]3 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]3 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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