Question

CD is the median of $\Delta$ABC. If height corresponding to AC in $\Delta$ADC is half of height corresponding to BD in $\Delta$BDC, then

• A. $\Delta$BDC is an isosceles triangle
• B. $\Delta$ABC is an isosceles triangle
• C. $\Delta$BDC is an equilateral triangle
• D. $\Delta$ABC is an equilateral triangle

Answer 1

The correct option is B

$\Delta$ABC is an isosceles triangle

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

Here, $AD=DB=\frac{AB}{2}$

Now, the area of triangle ABC = $\frac{1}{2}\times AB\times CF$

Area of triangle ADC = $\frac{1}{2}\times AD\times CF=\frac{1}{2}\times AC\times DE$ (Height corresponding to side AC is DE)

Area of Triangle CDB = $\frac{1}{2}\times BD\times CF$

Now since AD = DB, $\frac{1}{2}\times BD\times CF=\frac{1}{2}\times AD\times CF$

Thus, Area of $△ADC$ = Area of $△CDB$ = $\frac{1}{2}$Area of $△ABC$

Area of $△ADC$ = $\frac{1}{2}$Area of $△ABC$

$\Rightarrow \frac{1}{2}\times AC\times DE=\frac{1}{2}\times \frac{1}{2}\times AB\times CF$

Now given DE = $\frac{1}{2}CF$ (height corresponding to AC in $\Delta$ADC is half of height corresponding to BD in $\Delta$BDC)

$\Rightarrow \frac{1}{2}\times AC\times \frac{CF}{2}=\frac{1}{2}\times \frac{1}{2}\times AB\times CF$

Thus, AB = AC

Hence, ABC is an isosceles triangle.