Math, asked by aayuah41, 1 year ago

In the given figure ab=bc ac=cd,prove that angle bad: angle adb= 3:1


gaurav2013c: what is in the figure? Give the name of polygon

Answers

Answered by rohitkumargupta
45

HELLO DEAR,




In the diagram, we have two Isosceles triangles. For ΔABC, AB = BC and for ΔACD, AC = CD



In isosceles triangle, the two angles opposite to the equal sides are also equal.


So, for ΔABC, ∠BAC = ∠ACB and for ΔACD, ∠CAD = ∠ADC



As ∠ACB is outside angle of ΔACD ,



so ∠ACB = ∠CAD + ∠ADC



⇒ ∠ACB = 2× ∠ADC (As, ∠CAD = ∠ADC )



⇒ ∠BAC = 2× ∠ADC (As, ∠BAC = ∠ACB )



Now, according to the diagram,



∠BAD - ∠CAD = ∠BAC



⇒ ∠BAD - ∠ADC = 2× ∠ADC [As, ∠CAD = ∠ADC and ∠BAC = 2× ∠ADC]



⇒ ∠BAD = 3× ∠ADC



⇒ ∠BAD = 3× ∠ADB [As, ∠ADC and ∠ADB are same angles]



⇒∠BAD / ∠ADB = 3/1



so, ∠BAD : ∠ADB = 3 : 1





I HOPE ITS HELP YOU DEAR,


THANKS


Similar questions
Math, 1 year ago