Question

in the given figure AD is median. prove that ar(triangle ABD) = ar(triangle ACD)

Answer 1

Answer:

Proved!

Step-by-step explanation:

We know  that,

$\bigstar\; \boxed{\sf {Area\; of\; triangle= \dfrac{1}{2} \times Base \times Height\;\dfrac{1}{2} \times Base \times Height}}$

So,

$\it Area\; of\; triangle, ABD= \dfrac{1}{2} \times BD \times AN12\times BD \times AN$

And,

$\it Area\; of\;triangle\;ADC= \dfrac{1}{2} \times DC \times AN =12 \times BD \times AN\;\dfrac{1}{2} \times DC \times AN = 12 \times BD \timesAN$

(As we know BD  =  DC )

So,

Area of triangle ABD  = Area of triangle ADC                                  

(Hence proved)

$\rule{300}{1.5}$

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