Question

In fig d is a point on bc such that angle abd=anglecad ab=5cm ad=4cm ac=3cm

Answer 1

Answer:

Step-by-step explanation:

It is given that D is a point on BC such that ∠ABD=∠CAD.

Now, From ΔACD and ΔBCA, we have

∠ACD=BCA (Common angle)

∠ABD=∠CAD (Given)

By AA similarity, ΔACD is similar to ΔBCA.

Thus,

Now,

(Because of theorem of area of similar triangles)

Therefore, areaΔACD=16cm^{2}[/tex] and areaΔBCA=25cm^{2}[/tex]