Question

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

Answer 1

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.

Answer 2

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

$m∠ADB>m∠BDC.$