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In the given figure , AB = BC and AC= CD, prove that angle BAD : angle ADB = 3:1 |
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Answer» ANSWER: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] ⇒ (Prov
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