Thread: Deductive Geometry

Originally Posted by ose90

In the triangle ABC, the point P lies on the side AC such that angle BPC = angle ABC. Show that the triangles BPC and ABC are similar.
If AB = 4cm , Ac = 8cm , BP = 3cm , find the area of the triangle BPC.

Dear sir/madam,

I have done the part one, showing they are similar triangles.

Now solving the part 2.

Originally Posted by ose90

In the triangle ABC, the point P lies on the side AC such that angle BPC = angle ABC. Show that the triangles BPC and ABC are similar.
If AB = 4cm , Ac = 8cm , BP = 3cm , find the area of the triangle BPC.

The part one is because the two said triangles have two angles equal each to each that's why they are similar.

For part two, use their similar properties.

AB /AC = BP /BC
BC = (8*3)/4 = 6

AB /BC = BP /PC
PC = (6*3)/4 = 4.5

You have now the 3 sides of triangle BPC.
Use the Heron's formula for the area.

Originally Posted by ticbol

The part one is because the two said triangles have two angles equal each to each that's why they are similar.

For part two, use their similar properties.

AB /AC = BP /BC
BC = (8*3)/4 = 6

AB /BC = BP /PC
PC = (6*3)/4 = 4.5

You have now the 3 sides of triangle BPC.
Use the Heron's formula for the area.

Thanks, I have solved it on my own too.
Glad to know that I'm correct with my own genuine attempt.