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Question

CD is the median of ΔABC. If height corresponding to AC in ΔADC is half of height corresponding to BD in ΔBDC, then


A

ΔBDC is an isosceles triangle

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B

ΔABC is an isosceles triangle

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C

ΔBDC is an equilateral triangle

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D

ΔABC is an equilateral triangle

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Solution

The correct option is B

ΔABC is an isosceles triangle


Let, CF and DE are altituted o Δ ABC and Δ ADC.
The median bisects the side opposite to the vertex it is drawn from

Here, AD=DB=AB2

Now, the area of triangle ABC = 12×AB×CF

Area of triangle ADC = 12×AD×CF=12×AC×DE (Height corresponding to side AC is DE)

Area of Triangle CDB = 12×BD×CF

Now since AD = DB, 12×BD×CF=12×AD×CF

Thus, Area of ADC = Area of CDB = 12Area of ABC

Area of ADC = 12Area of ABC

12×AC×DE=12×12×AB×CF

Now given DE = 12CF (height corresponding to AC in ΔADC is half of height corresponding to BD in ΔBDC)

12×AC×CF2=12×12×AB×CF

Thus, AB = AC

Hence, ABC is an isosceles triangle.


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