MathDB

Let

ABC

be a right triangle with

\angle A=90^\circ.

The bisectors of the angles

B

and

C

meet each other in

I

and meet the sides

AC

and

AB

D

and

E

, respectively. Prove that

S_{BCDE}=2S_{BIC},

where

S

is the area function. [asy] import graph; size(200); real lsf = 0.5; pen dp = linewidth(0.7) + fontsize(10); defaultpen(dp); pen ds = black; pen ttqqcc = rgb(0.2,0,0.8); pen qqwuqq = rgb(0,0.39,0); pen xdxdff = rgb(0.49,0.49,1); pen fftttt = rgb(1,0.2,0.2); pen ccccff = rgb(0.8,0.8,1); draw((1.89,4.08)--(1.89,4.55)--(1.42,4.55)--(1.42,4.08)--cycle,qqwuqq); draw((1.42,4.08)--(7.42,4.1),ttqqcc+linewidth(1.6pt)); draw((1.4,10.08)--(1.42,4.08),ttqqcc+linewidth(1.6pt)); draw((1.4,10.08)--(7.42,4.1),ttqqcc+linewidth(1.6pt)); draw((1.4,10.08)--(4,4.09),fftttt+linewidth(1.2pt)); draw((7.42,4.1)--(1.41,6.24),fftttt+linewidth(1.2pt)); draw((1.41,6.24)--(4,4.09),ccccff+linetype("5pt 5pt")); dot((1.42,4.08),ds); label("

A

", (1.1,3.66),NE*lsf); dot((7.42,4.1),ds); label("

B

", (7.15,3.75),NE*lsf); dot((1.4,10.08),ds); label("

C

", (1.49,10.22),NE*lsf); dot((4,4.09),ds); label("

E

", (3.96,3.46),NE*lsf); dot((1.41,6.24),ds); label("

D

", (0.9,6.17),NE*lsf); dot((3.37,5.54),ds); label("

I

", (3.45,5.69),NE*lsf); clip((-6.47,-7.49)--(-6.47,11.47)--(16.06,11.47)--(16.06,-7.49)--cycle); [/asy]

1

Problems(2)