Question

In the figure, D is a point on BC, such that angle ABD= angle CAD. If AB=5cm , AD=4cm, AC=3cm. Find BC , DC , A(ACD):A(BCA)

Answer 1

Hey, I guess you forgot to attach the figure. Well, no worries in that.

Have a look at the attachment! ;)

In Δ ACD & Δ BCA,

∠ACD = ∠BCA ( Common )

∠CAD = ∠ABC ( Given )

The third angle is anyways equal, Using AA criteria.

Therefore, Δ ACD is similar to ΔBCA.

__________________

⟶ $\sf{\frac{AC}{BC}=\frac{AD}{BA}}$

⟶ $\sf{\frac{3}{BC}=\frac{4}{5}}$

⟶ 3 × 5 = 4 × BC

⟶ 15 = 4 × BC

⟶ $\boxed{\sf{BC=3.75}}$

Now using Theorem of Area of Similar Triangles,

$\sf{ \frac{area( \triangle \: {ACD})}{area( \triangle \: {BCA})} } = \frac{ {AD}^{2} }{ {BA}^{2} } = \frac{ {4}^{2} }{ {5}^{2} } = \frac{16}{25}$

_________________

Samajh aaya kuch? xD xD

Triangles isn't as hard as you think, just give it a try & you'll start lovin' it!

Have a great dayyyy! ;)