Question

From the following figure, we can say that $AB>CD$.

If true then enter $1$, else enter $0$.

Answer 1

Given, $\angle ABC=\angle ABD={70}^o$, $\angle ADC=\angle ADB={40}^o$ and $AB=AC$.

In $△ABC$,

since $AB=AC$, we get, $\angle ABC=\angle ACB={70}^o$ ...... [By Isosceles triangle property].

Also, $\angle ACB+\angle ACD={180}^o$ ......[Linear pairs]

$⟹$ ${70}^o+\angle ACD={180}^o$

$⟹$ $\angle ACD={180}^o-{70}^o={110}^o$

$⟹$ $\angle ACD={110}^o$.

Then, in $△ACD$,

$\angle ADC+\angle ACD+\angle CAD={180}^o$ ......[By angle sum property]

$⟹$ ${40}^o+{110}^o+\angle CAD={180}^o$

$⟹$ ${150}^o+\angle CAD={180}^o$

$⟹$ $\angle CAD={180}^o-{150}^o={30}^o$

$⟹$ $\angle CAD={30}^o$.

Since,in $△ACD$,

$\angle ADC>\angle CAD$ $⟹$ $AC>CD$.

$⟹$ $AB>CD$ ......[Since, given $AB=AC$].

Therefore, the statement is true and $1$ is the answer.