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

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 =
`(1)/(2)* base* height`

=
`(1)/(2)* AD* CD`

Area of ΔCDB =
`(1)/(2)* base* height`

=
`(1)/(2)* DB* CD`

Therefore ratio of the areas of ΔACD and ΔCDB is

i.e.
`(\Delta ACD)/(\Delta CDB)` =
`((1)/(2)* AD* CD)/((1)/(2)* DB* CD)`

=
`(AD)/(DB)`

∴ Area of the ratio of
`(\Delta ACD)/(\Delta CDB)` =
`(AD)/(DB)`

Hence proved