Question

In the given figure, $\angle BAD={65}^o,\angle ABD={70}^o,\angle BDC={45}^o$
(i) Prove that $AC$ is a diameter of the circle.
(ii) Find $\angle ACB$.

Answer 1

Given: $\angle BAD={65}^o,\angle ABD={70}^o,\angle BDC={45}^o$
(i) $\because$ ABCD is a cyclic quadrilateral.
In $\Delta$ ABD,
$\angle$ BDA $+\angle$ DAB $+\angle$ ABD$={180}^o$ By using sum property of ${\Delta }^s$
$\therefore$ $\angle$BDA $={180}^o-({65}^o+{70}^o)$
$={180}^o-{135}^o$
$={45}^o$
Now from $\Delta$ ACD,
$\angle$ADC $=\angle$ADB$+\angle$BDC
$={45}^o+{45}^o$ $(\because \angle$ BDA$=\angle$ADB$={45}^o)$
Hence, $\angle$D makes right angle belongs in semi-cricle therefore AC is a diameter of the circle.
(ii) $\angle$ACB$=\angle$ADB (Angles in the same segment of a circle)
$\therefore$ $\angle$ACB$={45}^o$ $(\because \angle$ ADB$={45}^o)$.