Question

In a figure D is point on BC such that angle ABC=angle CAD
if AB= 5 cm AD =4cm and AC =3cm Fing BC and A(triangle ACD):A(triangle BCA)

Answer 1

Answer:

Step-by-step explanation:

Answer 2

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, $\frac{AC}{BC}=\frac{AD}{BA}$

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

$15=4BC$

$BC=3.75$

Now, $\frac{a{\triangle}ACD}{a{\triangle}BCA}=\frac{AD^{2}}{BA^{2}}=\frac{4^{2}}{5^{2}}=\frac{16}{25}$

(Because of theorem of area of similar triangles)

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