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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 ΔABE ~ ΔCFB

**Solution:**

Let's draw a parallelogram ABCD as per the given question.

If two angles of one triangle are respectively equal to two angles of another triangle, then the two triangles are similar.

This is referred to as the AA similarity criterion for two triangles.

In ΔABE and ΔCFB,

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

∠AEB = ∠FBC [AE || BC and EB is a transversal, alternate interior angles]

Thus, ΔABE ~ ΔCFB (AA criterion)

**☛ Check: **NCERT Solutions Class 10 Maths Chapter 6

**Video Solution:**

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

NCERT Class 10 Maths Solutions Chapter 6 Exercise 6.3 Question 8

**Summary:**

If E is a point on the side AD produced of a parallelogram ABCD and BE intersect CD at F, we have proved that ΔABE ~ ΔCFB.

**☛ Related Questions:**

- In Fig. 6.39, ABC and AMP are two right triangles, right angled at B and M respectively. Prove that: (i) ΔABC ~ ΔAMP (ii) CA/PA = BC/MP.
- CD and GH are respectively the bisectors of ∠ACB and ∠EGF such that D and H lie on sides AB and FE of ∆ABC and ∆EFG respectively. If ∆ABC ~ ∆FEG, show that: (i) CD/GH =AC/FG (ii) ∆DCB ~ ∆HGE (iii) ∆DCA ~ ∆HGF
- In Fig. 6.40, E is a point on side CB produced of an isosceles triangle ABC with AB = AC. If AD ⊥ BC and EF ⊥AC, prove that ∆ ABD ~ ∆ ECF.
- Sides AB and BC and median AD of a triangle ABC are respectively proportional to sides PQ and QR and median PM of ∆ PQR (see Fig. 6.41). Show that ∆ ABC ~ ∆ PQR.

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