This is the basic and conceptual question from similarity chapter in which we are to

(i) Prove that △ACD is similar to △BCA

(ii) Find BC and CD

(iii) Find the area of △ACD : area of △ABC.

We have given a figure of triangle in which ∠ABC=∠DAC and AB=8cm,AC=4cm,AD=5cm

## From the question it is given that,

∠ABC=∠DAC

AB=8cm,AC=4cm,AD=5cm

(i) Now, consider △ACD and △BCA

∠C=∠C … [common angle for both triangles]

∠ABC=∠CAD … [from the question]

So, △ACD∼△BCA … [by AA axiom]

(ii) AC/BC=CD/CA=AD/AB

Consider AC/BC=AD/AB

4/BC=5/8

BC=(4×8)/5

BC=32/5

BC=6.4cm

Then, consider CD/CA=AD/AB

CD/4=5/8

CD=(4×5)/8

CD=20/8

CD=2.5cm

(iii) from (i) we proved that, △ACD∼△BCA

area of △ACB/area of △BCA=AC2/AB2

=42/82

=16/64

By dividing both numerator and denominator by 16, we get,

=41

Therefore, the area of △ACD : area of △ABC is 1:4.