Question

In the adjoining figure, D is a point on BC such that ∠ABD = ∠CAD. If AB = 5cm, AC = 3cm and AD

=4cm, find
i. BC

ii. DC

iii. area of △ACD : area of △BCA.​

Answer 1

$\huge\star\underline\mathfrak\pink{Answer:-}$

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,

\frac{3}{BC}=\frac{4}{5}

BC=3.75

(Because of theorem of area of similar triangles)

Now,

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