Question

14
(a) In the figure (1) given below, CDE is an equilateral triangle fontes
CD of a square ABCD Show that AADE - ABCE and hence, AEB.
triangle
(b) In the figure (11) given below, O is a point in the interior of a quase AB,
that OAB is an equilateral triangle. Show that OCD is an ismele
A
C
E
(1)​

Answer 1

a-According to figure,

∠ADC=∠BCD=90°→square

∠CDE=∠DCE=60°→equilateral_triangle

∴∠ADE=∠BCE=150°

InΔADEandΔBCE

AD=BC

DE=CE

∠ADE=∠BCE

ΔADE≅ΔBCE

b-ΔOAB is equilateral triangle then

ln ΔAOD and ΔBOC

AD=BC (sides of the square)

∠DAO=∠CBD=30°(90°−angle of equilateral Δ(60°))

AO=OB (sides of equilateral of triangle)

ΔAOD≅ΔBOC (SAS criterion)

then OD=OC

So ΔCOD is an isosceles triangle