Question

In Fig.∠A = ∠CED, prove that ΔCAB~ ΔCED. Also, find the value of x.

Answer 1

We have,

∠A = ∠CED

In ΔCAB and ΔCED

∠C = ∠C(common)

∠A = ∠CED(Given)

Then, ΔCAB ~ ΔCED (By AA similarity)

So,\(\frac{CA}{CE}\) =\(\frac{AB}{ED}\)  (Corresponding parts of similar triangle area proportion)

Or,15/9 = 9/x 

Or, 15x = 90 

Or, x = 90/6 

Or, x = 6cm.