Q24 and Q26 with proper equation Q24 In a ΔABC, the internal bisector of angle A meets opposite - Maths - Triangles

Question

Q24 and Q26 with proper equation

Q24 In a ΔABC, the internal bisector of angle A meets
opposite side BC at point D Through vertex C, line CE is drawn
parallel to DA which meets BA produced at point E Show that
ΔACE is isosceles

Q26 In ΔABC, D is a point on BC such that AB = AD = BD = DC
Show that: ∠ADC : ∠C = 4 : 1

Sushant Ranjan · Accepted Answer

Dear Student,

Solution 26:
<img alt="" src=
"https://s3mn%20mnimgs%20com/img/shared/content_ck_images/images/Untitled(136)%20png">
Here AB,AD and BD are side of a triangle and all the sides are
equal as per question
This means that triangle is equilateral triangle and each angle is
of 60 deg
Now ∠ADC + ∠ADB = 180°∠ADB= 60° as the triangle is equilateral triangle 
⇒ ∠ADC =180° - 60° = 120°
1Now in △ADC , AD= DC{ given }So the triangle is isosceles triangle 
 i e∠DAC =∠ACD 
2Let the angle be y°
Now we know that sum of all the angle of triangle is 180°
or,∠ACD+∠ADC+∠DAC =180°or,∠ADC+2y=180°or,120° + 2y =180° {by eq 1 and 2}⇒2y=60°⇒y=30°Now ∠ADC : ∠C = 120°:30° = 4:1 Hence proved 
Hope this would have cleared your doubt

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