In a right-angled triangle ABC, the height CH, drawn from the top of the right angle C, divides the hypotenuse AB

In a right-angled triangle ABC, the height CH, drawn from the top of the right angle C, divides the hypotenuse AB into two segments AH = 5cm and CH = 4cm. Search: leg BC

Consider a triangle ABC.

Since triangle ABC and C = 90 °,

CBA = 90 ° – CAB.

Consider triangles ACH and BCH.

Since CH is the height of triangle ABC, CH is perpendicular to AB.

Therefore, the angles CHA = CHB = 90 ° and the triangles ACH and BCH are rectangular.

notice, that

ACH = 90 ° – CAB = CBA,

BCH = 90 ° – CBA = CAB.

Therefore, triangles ACH and BCH are similar in three angles. Then we have:

CH / AH = BH / CH,

CH ^ 2 = AH * BH = 5 * 4,

CH = 2 * √5.

By the Pythagorean theorem, from the triangle BCH we have:

BC ^ 2 = CH ^ 2 + BH ^ 2 = 5 * 4 + 4 ^ 2 = 36,

BC = 6.

Answer: 6.