>
#mathematics
,
#triangles
,
#geometry
>
PQRS is a square. The diagonals intersect at E. In the middle of line PS,there are diagonals A and B which also intersect at E,forming 3 triangles in triangle PES,and PA=PS?
Anonymous
more than 4 weeks ago
PQRS is a square. The diagonals intersect at E. In the middle of line PS,there are diagonals A and B which also intersect at E,forming 3 triangles in triangle PES,and PA=PS?
Prove that triangle AEB is an isosceles triangle
2 views
2 views
2 answers
Sort By
ChatGPT
more than 4 weeks ago
No, it is not possible for PA to be equal to PS in triangle PES. Since A is the midpoint of PS, PA must be half the length of PS. Therefore, PA is not equal to PS.
0
Google Gemini AI
more than 4 weeks ago
In the figure, PQRS is a square and diagonals PR and QS intersect at E. Let PA = PS = x.
Since PQRS is a square, PR = QS = √2x (using the diagonal of a square theorem)
In triangle PES, we have:
PE = PS - ES = x - √2x/2 = √2x/2
Similarly, SE = √2x/2
Now, in triangle PAE, we have:
AE = √(AP^2 + PE^2) = √(x^2 + (√2x/2)^2) = √(x^2 + x^2/2) = √(3x^2/2) = √(3x/2)
Therefore, AE = √(3x/2).
Since PQRS is a square, PR = QS = √2x (using the diagonal of a square theorem)
In triangle PES, we have:
PE = PS - ES = x - √2x/2 = √2x/2
Similarly, SE = √2x/2
Now, in triangle PAE, we have:
AE = √(AP^2 + PE^2) = √(x^2 + (√2x/2)^2) = √(x^2 + x^2/2) = √(3x^2/2) = √(3x/2)
Therefore, AE = √(3x/2).
0
