Question

E is a point on side ad produced of a parallelogram ABCD and be intersects CD at F. Prove that ∆ABE ~ ∆CFB.​

Answer 1

Answer:

In a parallelogram opposite angles are equal to each other. Then find the relation between angles of both triangle's to show them similar.

If two angles of a triangle are respectively equal to two angles of another triangle, then Triangles are similar (because by the angle sum property of a triangle their third angle will also be equal ) and it is called AA similarity criterion.

In ΔABE and ΔCFB,

∠A=∠C                (Opposite angles of a parallelogram)

∠AEB=∠CBF   (Alternate interior angles as AE || BC)

∴ ΔABE ΔCFB   (By AA similarity criterion)

Step-by-step explanation:

Answer 2

Given

• ABCD is a parallelogram.
• BE intersects CD at F.

To prove

• ∆ABE ∼ ∆CFB

Solution

$\sf{⟶}$ In ∆ABE and ∆CFB, we have

→ ∠AEB = ∠CBF⠀⠀⠀⠀[ᴀʟᴛᴇʀɴᴀᴛᴇ ⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀ᴀɴɢʟᴇs]

→ ∠A = ∠C⠀⠀⠀⠀[ᴏᴘᴘᴏsɪᴛᴇ ᴀɴɢʟᴇs ᴏғ ᴀ ⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀ᴘᴀʀᴀʟʟᴇʟᴏɢʀᴀᴍ]

⠀

$\tt\longmapsto{}$ Thus, by AA criterion of similarity, we have

∆ABE ∼ ∆CFB

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