PS ≅ FK, RF ≅ GP ∠R ≅ ∠G
Solution:
Given in ΔKFR, KF = 10, FR = 16, KR = 20
In ΔGPS, GP = 8, GS = 10, PS = 5
In the given image the two triangles are similar.
That is, ΔKFR ≅ ΔGPS
We know that the corresponding parts of the similar triangles are congruent.
(1) ∠F ≅ ∠P (given)
So, PS ≅ FK
(2) ∠R ≅ ∠G (given)
∠F ≅ ∠P (given)
So, RF ≅ GP
(3) ∠R ≅ ∠G (given in the image)
Hence PS ≅ FK, RF ≅ GP ∠R ≅ ∠G.
