E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Show that ΔABE ~ ΔCFB
Let's draw a parallelogram ABCD as per the given question.
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)
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
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.
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