# The height BH is drawn in an acute-angled triangle ABC. Perpendiculars hk and hm were dropped from point H

**The height BH is drawn in an acute-angled triangle ABC. Perpendiculars hk and hm were dropped from point H to side AB and BC, respectively. Prove that triangle mbk is similar to triangle ABC**

To prove this statement, we use the following criteria for the similarity of triangles:

If two angles of one triangle are respectively equal to two angles of another, then such triangles are similar.

If the two sides of one triangle are proportional to the two sides of the other triangle and the angles formed by these sides are equal, then such triangles are similar.

Triangle ABH is similar to triangle BHK (in two corners).

Therefore, the following proportion is true:

│BK│ / │BK│ = │BH│ / │AB│;

│AB│ * │BK│ = │BH│².

Likewise, triangle CBH is similar to triangle BHM:

| СB | * | BM | = | BH | ²;

| AB | * | BK | = | С | * | BM |.

| AB | / | BM | = | СB | / | BK |, side AB is similar to side BM, side CB is similar to side BK, angle B is common for both triangles, which means that triangle MBK is similar to ABC in two sides and the angle between them.

Q.E.D.