Question

Suppose that triangle ABC is a right triangle with the right angle at C. Let line segment CD be the perpendicular from point C to the hypotenuse line segment AB. Show that the ratio of the areas of triangles ADC and DCB is the same as the ratio AD: DB

Answer 1

Step-by-step explanation:

Given:

Let triangle  ACD is aright angle triangle with right angle at C. A line perpendicular to AB join C.

Therefore we can say that line segment CD divides angle at C into two equal angles.

So in ΔACD and ΔCDB

           ∠ ACD = ∠DCB

and  ∠ADC = ∠BDC = 90°

and   CD =CD

∴ we can say that ΔACD and ΔCDB are similar triangles.

∴ Area of ΔACD = $\frac{1}{2}\times base\times height$

                           = $\frac{1}{2}\times AD\times CD$

Area of ΔCDB = $\frac{1}{2}\times base\times height$

                           = $\frac{1}{2}\times DB\times CD$

Therefore ratio of the areas of  ΔACD and ΔCDB is

i.e. $\frac{\Delta ACD}{\Delta CDB}$ = $\frac{\frac{1}{2}\times AD\times CD}{\frac{1}{2}\times DB\times CD}$

                = $\frac{AD}{DB}$

∴ Area of the ratio of $\frac{\Delta ACD}{\Delta CDB}$ =  $\frac{AD}{DB}$

Hence proved