e is a point on the side ad produced of a parallelogram

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

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Question

E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Show that △ABE∼△CFB.

Medium Open in App

Updated on : 2022-09-05

In △ABE and △CFB,

Solution Verified by Toppr

∠ABE=∠CFB (Alternate angles)

∠BAE=∠BCF (opposite angles of a parallelogram)

∴ By AA criterion of similarity, △ABE ∼ △CFB

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Question 8 E is a point on the side AD produced of a parallelogram ABCD and BE intersect CD at F. Show that Δ ABE ∼Δ CF B.

Question 8 E is a point on the side AD produced of a parallelogram ABCD and BE intersect CD at F. Show that Δ ABE ∼Δ CF B.

Byju's Answer Standard X Mathematics AAA Similarity Question 8 E ... Question Question 8

E is a point on the side AD produced of a parallelogram ABCD and BE intersect CD at F. Show that

Δ A B E ∼ Δ C F B . Open in App Solution

In Δ A B E and Δ C F B , ∠ A = ∠ C

(Opposite angles of a parallelogram)

∠ A E B = ∠ C B F

(Alternate interior angles as AE || BC)

∴ Δ A B E ∼ Δ C F B

(By AAA similarity criterion)

Suggest Corrections 13 Video Solution

NCERT - Grade 10 - Mathematics - Triangles - Q21

MATHEMATICS

07:37 Min | 72 Views

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SIMILAR QUESTIONS

Q.

E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Show that ΔABE ∼ ΔCFB

Q. In the figure given below,

E

is a point on the side

A D

produced of a parallelogram

A B C D and B E intersects C D at F . Show that △ A B E ∼ △ C F B .

Q.

State True or False-

E is a point on the side produced of a parallelogram and intersects at F.

△ A B E is similar to △ C F B :

Q. Point E bisects side CD of a parallelogram ABCD and CF intersects DA produced at G such that

C F | | A E

, where F is a point on AB. The value of

A D × G F – C F × A G is

Q.

A B C D

is a parallelogram. The circle through

A , B and C intersect C D

(produced if necessary) at

E . Prove that A E = A D . View More EXPLORE MORE AAA Similarity

Standard X Mathematics

स्रोत : byjus.com

Ex 6.3, 8

Ex 6.3, 8 E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Show that ΔABE ∼ ΔCFB Given: A parallelogram ABCD where E is point on side AD produced & BE intersects CD at F To Prove: ΔABE ∼ ΔCFB. Proof: In parallelogram ABCD , opposite angles are

Check sibling questions

Ex 6.3, 8 - Chapter 6 Class 10 Triangles (Term 1)

Last updated at March 16, 2023 by Teachoo

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Transcript

Ex 6.3, 8 E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Show that ΔABE ∼ ΔCFB Given: A parallelogram ABCD where E is point on side AD produced & BE intersects CD at F To Prove: ΔABE ∼ ΔCFB. Proof: In parallelogram ABCD , opposite angles are equal, Hence, ∠A = ∠C Also, In parallelogram ABCD opposite sides are parallel, AD ∥ BC Now since AE is AD extended, AE ∥ BC and BE is the traversal ∴ ∠ AEB = ∠ CBF Now in Δ ABE & Δ CFB ∠A = ∠C ∠ AEB = ∠ CBF ∴ Δ ABE ∼ Δ CFB Hence proved

Next: Ex 6.3, 9 →

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.

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