Question

1) In the figure, triangle ABC is an isosceles triangle such that AB = AC. P is any point on AC. Through point C, a line is drawn to intersect BP produced at Q. Such that angle ABQE is congruent to angle ACQ Prove that: angle AQC = 90° + 1/2 angle BAC. ​

Answer 1

$\large\underline{\sf{Solution-}}$

Given that In $\triangle\: ABC$

AB = AC

$\rm \implies \angle\:ABC = \angle\:ACB$

{ $\because$Angle opposite to equal sides are equal}

Let assume that

$\angle\:ABC = \angle\:ACB = x \: (say)$

Now, In $\triangle\: ABC$

$\rm \: \angle \: BAC + x + x = 180 \degree \:$

$\rm \: \angle \: BAC + 2x = 180 \degree \:$

$\rm\implies \: \angle \: BAC = 180 \degree \: - \: 2x$

Now,

$\rm\implies \: \angle\:BQC = \angle \: BAC = 180 \degree \: - \: 2x - - - (1)$

{ $\because$Angle in same segments are equal }

Also,

$\rm \: \angle\:AQB = \angle\:ACB = x - - - (2)$

{ $\because$Angle in same segments are equal }

Now, As

$\rm \: \angle \: BAC = 180 \degree \: - \: 2x$

$\rm \: 2x = 180 \degree \: - \: \angle\:BAC$

$\rm\implies \:x = 90 \degree \: - \: \frac{1}{2} \: \angle\:BAC - - - (3)$

Now, Consider

$\rm \: \angle\:AQC$

$\rm \: = \: \angle\:AQB + \angle\:CQB$

$\rm \: = \: x + \angle\:BAC$

$\rm \: = \: 90\degree \: - \frac{1}{2}\angle\:BAC + \angle\:BAC$

$\rm \: = \: 90\degree \: + \frac{1}{2}\angle\:BAC$

Hence,

$\bf\implies \:\angle\:AQC = \: 90\degree \: + \frac{1}{2}\angle\:BAC$

Hence, Proved

$\rule{190pt}{2pt}$

Additional Information :-

1. Angle in semi-circle is right angle.

2. Angle subtended at the centre of a circle by an arc is double the angle subtended on the circumference of a circle by the same arc.

3. Sum of the opposite angles of a cyclic quadrilateral is supplementary.

4. Exterior angle of a cyclic quadrilateral is equals to interior opposite angle.

Answer 2

Answer:

Here you go.

Step-by-step explanation: