E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Show that ΔABE ~ ΔCFB.
Given: ABCD is a parallelogram, where BE intersects CD at F. E is a point on the side AD produced.
To Prove: ΔABE ~ ΔCFB

Proof: In ∆ABE and ∆CFB,
\(\angle\)A = \(\angle\)C (Opposite angles of a parallelogram)
\(\angle\)AEB = \(\angle\)CBF (Alternate interior angles as AE || BC)
∴ ∆ABE ∼ ∆CFB (By AA similarity criterion)
Hence Proved
In the adjoining figure, \(PQ \parallel XY \parallel BC\), \(AP=2\ \text{cm}, PX=1.5\ \text{cm}, BX=4\ \text{cm}\). If \(QY=0.75\ \text{cm}\), then \(AQ+CY =\)
In the adjoining figure, \( \triangle CAB \) is a right triangle, right angled at A and \( AD \perp BC \). Prove that \( \triangle ADB \sim \triangle CDA \). Further, if \( BC = 10 \text{ cm} \) and \( CD = 2 \text{ cm} \), find the length of } \( AD \).
If a line drawn parallel to one side of a triangle intersecting the other two sides in distinct points divides the two sides in the same ratio, then it is parallel to the third side. State and prove the converse of the above statement.
Leaves of the sensitive plant move very quickly in response to ‘touch’. How is this stimulus of touch communicated and explain how the movement takes place?
Read the following sources of loan carefully and choose the correct option related to formal sources of credit:
(i) Commercial Bank
(ii) Landlords
(iii) Government
(iv) Money Lende