ABCD is a parallelogram point E is on side BC. Line DE intersects ray AB in point T. Prove that DE ×BE= CE × TE
Answers
Answer:
By BPT Theorem we can prove this sum
Answer:
DE × BE = CE × TE
Step-by-step-explanation:
NOTE: Refer to the attachment for the diagram.
We have given that,
□ABCD is a parallelogram.
∴ Seg AB ∥ seg CD - - [ Opposite sides of a parallelogram ]
∴ Seg AT ∥ Seg CD, - - [ A - B - T ]
Now, Seg AT ∥ Seg CD and line TD is transversal.
∴ ∠ATD ≅ ∠CTD - - [ Alternate angles ]
⇒ ∠BTE ≅ ∠CDE - - - ( 1 ) [ A - B - T, T - E - D ]
Now,
In △BET & △CED,
∠BTE ≅ ∠CDE - - [ From ( 1 ) ]
∠BET ≅ ∠CED - - [ Vertically opposite angles ]
∴ △BET ∼ △CED - - [ AA test of similarity ]
⇒ BE / CE = TE / DE - - - [ c. s. s. t. ]
∴ BE × DE = CE × TE
i. e.
DE × BE = CE × TE
Hence proved!
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Additional Information:
1. Similar triangles:
Two triangles are similar if their corresponding angles are congruent and the corresponding side are in proportion.
2. Area of two similar triangles:
If two triangles are similar, then the ratio of areas of those triangles is equal to the ratio of squares of the corresponding sides.
3. Tests of similarity of triangles:
A. AAA test
B. AA test
C. SAS test
D. SSS test
4. AAA test:
In two triangles, if the all corresponding angles are congruent, then the two triangles are similar.
5. AA test:
In two triangles, if only two corresponding angles are congruent, then the two triangles are similar.
6. SAS test:
In two triangles, if two corresponding sides are in proportion and the corresponding angle made by them is congruent, then two triangles are similar.
7. SSS test:
In two triangles, if all three corresponding sides are in proportion, then the two triangles are similar.
