Question

$\angle APB=x,\angle AQB=y$ and $\angle AOB=z$
Then prove that: $\angle x+\angle y=\angle z$
Also, find the value of angleCAB

Answer 1

Given: $\angle APB=x,\angle AQB=y$ and $\angle AOB=z$
To prove: $\angle x+\angle y=\angle z$
As shown in figure, Let $\angle ACB=c$ and $\angle ADB=d$
$\therefore \angle QCP=({180}^{\circ }-c)$ and $\angle QDP=({180}^{\circ }-d)$
So, in quadrilateral, PCQD
$\angle CPD+\angle PDQ+\angle DQC+\angle QCP={360}^{\circ }$
$x+({180}^{\circ }-d+y({180}^{\circ }-c\left)\right)={360}^{\circ }$
(As we know, sum of all angle of quadrilateral is ${360}^{\circ }$)
$x+y={360}^{\circ }-{360}^{\circ }+c+d$
$x+y=c+d$
We know that, $\angle AOB=2\angle ACB$
(By the theorem, the angle formed at the centre of the circle by the lines originating form two points on the circle’s circumference is double the angle formed on the circumference of the circle by lines originating form the same points)
$\because \angle AOB=2\angle ACB$
$\Rightarrow z=2c\left(2\right)$
$\because \angle AOB=2\angle ADB$
$\Rightarrow z=2d\left(3\right)$
By adding (2) and (3),
$2c+2d=2z$
$\Rightarrow z=c+d$ ..... (4)
From equation (1) and (4),
Henced proved, $\angle x+\angle y=\angle z$
Here BD =OD
And we know OD =OB (Radius of circle)
So, In $\Delta OBD$, We know OD = BD = OB, so this is a equilateral triangle, So all internal angle $\Delta OBD$ are at ${60}^{\circ }$, So
$\angle OBD=\angle ODB=\angle BOD={60}^{\circ }$
And also given : CD perpendicular to AB, So from this information we know in equilateral triangle altitude drawn from any vertex bisect the vertex angle and also bisects the opposite side.
$\angle EDB=\angle EDO={30}^{\circ }$ ....... (1)
And OE =BE.
And we also know if a perpendicular draw from centre (OE) to any chord (CD), so, that bisect chord, So
CE =DE ………….(2)
Now in $\Delta BED$ and $\Delta BEC$,
CE =DE (From equation 1)
And
$\angle BCE$ and $\angle BCD$ are same angles, So
$\angle BCD={30}^{\circ }$
In $\Delta BCE$ from angle sum property we get
$\angle BCE+\angle BEC+\angle CBE={180}^{\circ }$
$\angle CBE={60}^{\circ }$
And We know “The angle inscribed in a semicircle is always a right angle”, So
$\angle ACB={90}^{\circ }$
In $\Delta ABC$ from angle sum property we get
$\angle CBA+\angle ACB+\angle CAB={180}^{\circ }$, Substitute all values, we get
${60}^{\circ }+{90}^{\circ }+\angle CAB={180}^{\circ }$ (As $\angle CBA$ and $\angle CBE$ are same angle)
$\angle CAB={30}^{\circ }$