Question

E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F . Show that Triangle ABE similar to Triangle CFB​

Answer 1

$\huge\sf\blue{Given}$

✭ ABCD is a parallelogram

✭ AD is extended to E

✭ Be intersects CD at F

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$\huge\sf\gray{To\;Prove}$

➢ ∆ CFB ~ ∆ ABE

$\rule{110}1$

$\huge\sf\purple{Steps}$

In ∆ CFB and ∆ ABE

$\sf\dashrightarrow\angle ABE = \angle DCB$ (opp angles of llgm)

$\sf\dashrightarrow\angle AEB = \angle CBF$ (BC || AD and AD is produced to E and BE is transversal)

$\therefore$ ∆ CFB ~ ∆ ABE (AA criteria)

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$\large\sf{\underline{\bullet{ More\: about \: Similarly}}}$

➳ When a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points,the other two sides are divided in the same ratio

➳ We can even prove the Pythagoras theorem using this concept

➳ There are 4 criteria to prove similarity

◕ AA criteria

◕ AAA criteria

◕ SAS criteria

◕ SSS criteria

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Answer 2

Answer:

Step-by-step explanation:

Given,

ABCD is parallelogram.

To Prove,

ABE ~ CFB​

Solution,

In ΔABE and ΔCFB,

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

∠AEB = ∠CBF (Alternate interior angles)

ΔABE ~ ΔCFB (By AA similarity criterion)

Hence Proved.

Extra Information :-

AA similarity criterion - If two angles of one triangle are equal to two angles of another triangles, then the two triangles are similar.