◻ABCD is a parallelogram point E is on side BC. Line DE intersects ray AB in point T. Prove that DE × BE = CE × TE.
Practice Set 1.3 | Q 7 | Page 22
◻ABCD is a parallelogram point E is on side BC. Line DE intersects ray AB in point T. Prove that DE × BE = CE × TE.
Given: ◻ABCD is a parallelogram
To prove: DE × BE = CE × TE
Proof: In ∆BET and ∆CED
∠BET = ∠CED (Vertically opposite angles)
∠BTE = ∠CDE (Alternate angles, AT || CD and DT is a transversal line)
By AA test of similarity
∆BET ∼ ∆CED
\[\therefore \frac{BE}{CE} = \frac{ET}{ED} \left( \text{ Corresponding sides are proportional } \right)\]
\[ \Rightarrow BE \times ED = CE \times ET\]
Answer:-
Given: ◻ABCD is a parallelogram and E is a point on side BC. Line DE intersects ray AB at point T.
To prove: DE × BE = CE × TE.
Proof:
We will use the concept of similarity of triangles.
Consider ∆BET and ∆CED.
By the construction of parallelogram ABCD, we know that AB || CD and BC || AD.
Thus, by the alternate angles theorem, we have ∠BTE = ∠CDE and ∠BET = ∠CED.
Also, as ABCD is a parallelogram, we have AB = CD and AD = BC.
Now, by the AA (angle-angle) test of similarity, we can conclude that ∆BET ~ ∆CED.
Therefore, we have BE/CE = ET/ED (corresponding sides are proportional).
Multiplying both sides by DE, we get:
BE × ED = CE × ET
which is the required result. Hence, the proof is complete.
Chapter 1. Similarity- Practice Set 1.3 – Page 22
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