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
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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
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